💡Alright, so I have this problem here where I need to use the method of contradiction to prove that given real numbers ( a ) and ( b ), the quadratic equation ( x^2 + ax + b = 0 ) has at least one real root. The question is asking me what assumption I need to make to start this proof by contradiction. The options are:A: The equation ( x^2 + ax + b = 0 ) has no real roots.B: The equation ( x^2 + ax + b = 0 ) has at most one real root.C: The equation ( x^2 + ax + b = 0 ) has at most two real roots.D: The equation ( x^2 + ax + b = 0 ) has exactly two real roots.Okay, so first, let me recall what proof by contradiction is. It's a method where you assume the opposite of what you want to prove and then show that this assumption leads to a contradiction. Therefore, the original statement must be true.In this case, the statement to prove is: "Given that ( a ) and ( b ) are real numbers, then the equation ( x^2 + ax + b = 0 ) has at least one real root."So, the negation of this statement would be: "Given that ( a ) and ( b ) are real numbers, the equation ( x^2 + ax + b = 0 ) has no real roots."Looking at the options, that's exactly what option A says. So, if I assume option A, that the equation has no real roots, and then show that this leads to a contradiction, then the original statement must be true.But wait, let me make sure I'm not missing something. The other options are about having at most one or two roots, or exactly two roots. But the original statement is about having at least one real root. So, the negation should be that it has fewer than one real root, which is zero real roots. That's exactly what option A is.Option B says "at most one real root." That's not the negation of "at least one real root." Because "at most one" could still include the possibility of having one real root, which doesn't contradict the original statement. So, assuming B wouldn't necessarily lead to a contradiction because it's not the direct opposite.Option C says "at most two real roots." But a quadratic equation can have at most two real roots, so assuming that doesn't really help in contradiction because it's always true. So, that's not useful.Option D says "exactly two real roots." But the original statement is about having at least one, so assuming exactly two might not necessarily lead to a contradiction either. Plus, the negation isn't about the number of roots being two, but rather about having none.So, going back, the correct assumption for contradiction should be the direct negation of the original statement, which is option A: the equation has no real roots.To further solidify this, let me think about how the proof would go. If I assume that the equation ( x^2 + ax + b = 0 ) has no real roots, then the discriminant must be negative. The discriminant ( D ) of a quadratic equation ( ax^2 + bx + c = 0 ) is ( D = b^2 - 4ac ). In our case, the equation is ( x^2 + ax + b = 0 ), so ( a = 1 ), ( b = a ), and ( c = b ). Therefore, the discriminant is ( D = a^2 - 4b ).If there are no real roots, then ( D < 0 ), so ( a^2 - 4b < 0 ), which implies ( a^2 < 4b ). Now, if I can show that this leads to a contradiction, then the original assumption must be false, meaning the equation does have at least one real root.But wait, how would I show that ( a^2 < 4b ) leads to a contradiction? Maybe I can use some properties of real numbers or another theorem. Hmm, perhaps I can consider the function ( f(x) = x^2 + ax + b ). If the equation has no real roots, then the function never crosses the x-axis, meaning it's always positive or always negative. Since the coefficient of ( x^2 ) is positive (1), the parabola opens upwards, so it would be always positive if there are no real roots. Therefore, ( f(x) > 0 ) for all real ( x ).But then, if ( f(x) = x^2 + ax + b > 0 ) for all ( x ), we can complete the square to find the minimum value of the function. Completing the square:( f(x) = x^2 + ax + b = (x + frac{a}{2})^2 - frac{a^2}{4} + b ).The minimum value occurs at ( x = -frac{a}{2} ), and the minimum value is ( -frac{a^2}{4} + b ). Since ( f(x) > 0 ) for all ( x ), this minimum must be greater than 0:( -frac{a^2}{4} + b > 0 )Which simplifies to:( b > frac{a^2}{4} )But earlier, from the discriminant, we had ( a^2 < 4b ), which is the same as ( b > frac{a^2}{4} ). So, this doesn't lead to a contradiction; it's consistent.Hmm, so maybe my initial thought process is missing something. If assuming no real roots doesn't lead to a contradiction, then perhaps the original statement isn't always true? But wait, that can't be right because a quadratic equation with real coefficients must have either one or two real roots, or two complex roots. But complex roots come in pairs, so if it has no real roots, it has two complex roots.Wait, but the original statement says "has at least one real root." So, if it has two complex roots, then it doesn't have any real roots, which would make the original statement false. But that contradicts what I know about quadratic equations. Wait, no, actually, a quadratic equation with real coefficients can have either two real roots or two complex conjugate roots. So, it's possible for it to have no real roots, which would mean the original statement is not always true.But the problem says "Given that ( a ) and ( b ) are real numbers, then the equation ( x^2 + ax + b = 0 ) has at least one real root." So, is this statement actually true? Because if the discriminant is negative, it has no real roots.Wait, maybe I misread the problem. Let me check again. It says "Given that ( a ) and ( b ) are real numbers, then the equation ( x^2 + ax + b = 0 ) has at least one real root." So, is this always true? No, because as I just thought, if the discriminant is negative, it has no real roots.So, does that mean the original statement is false? But the problem is asking to prove it using contradiction, which suggests that the statement is true. So, perhaps there's a misunderstanding here.Wait, maybe the problem is misstated. Or perhaps I need to consider something else. Let me think again.If ( a ) and ( b ) are real numbers, then the quadratic equation ( x^2 + ax + b = 0 ) can have either two real roots, one real root (if discriminant is zero), or two complex roots (if discriminant is negative). So, it's not necessarily true that it has at least one real root. Therefore, the original statement seems to be false.But the problem is asking to prove it using contradiction, which suggests that the statement is true. So, perhaps I'm missing something.Wait, maybe the problem is in the context of real numbers, but perhaps it's considering multiplicity? Like, counting real roots with multiplicity? But even so, if the discriminant is negative, there are no real roots, regardless of multiplicity.Alternatively, maybe the problem is assuming that ( a ) and ( b ) are such that the equation has at least one real root, but that's not what's given. The given is just that ( a ) and ( b ) are real numbers.Wait, unless there's a typo in the problem. Maybe it's supposed to say "has at most one real root" or something else. But as it stands, the statement is not universally true.But since the problem is asking to prove it using contradiction, perhaps I need to proceed under the assumption that the statement is true, even though in reality, it's not. Or maybe I'm misapplying something.Alternatively, perhaps the problem is referring to real roots in a different sense, like considering repeated roots as one root, but even then, if the discriminant is negative, there are no real roots.Wait, maybe I need to think differently. Perhaps the problem is not about quadratic equations in general, but about something else. Let me reread the problem."Given that ( a ) and ( b ) are real numbers, then the equation ( x^2 + ax + b = 0 ) has at least one real root."So, it's a quadratic equation with real coefficients, and we need to prove it has at least one real root. But as I know, quadratics can have zero, one, or two real roots. So, the statement is not always true.Therefore, perhaps the problem is incorrectly stated, or maybe I'm misinterpreting it. Alternatively, maybe the problem is in a context where ( a ) and ( b ) are such that the equation must have at least one real root, but that's not specified.Alternatively, perhaps the problem is to show that it has at least one real root under certain conditions, but as given, it's just real coefficients.Wait, maybe I need to consider the function ( f(x) = x^2 + ax + b ) and use the Intermediate Value Theorem. Since it's a continuous function, and as ( x ) approaches infinity, ( f(x) ) approaches positive infinity, and as ( x ) approaches negative infinity, ( f(x) ) also approaches positive infinity (since the leading coefficient is positive). Therefore, if the function ever dips below the x-axis, it must cross it at least twice. But if it never dips below, then it has no real roots.But wait, if the function has a minimum value above the x-axis, then it has no real roots. So, again, it's possible for the equation to have no real roots.Therefore, the original statement is not always true, which contradicts the idea of proving it via contradiction. So, perhaps the problem is misstated.Alternatively, maybe the problem is to show that it has at least one real root under certain conditions, like if ( b ) is negative or something. But as given, it's just real ( a ) and ( b ).Wait, maybe I'm overcomplicating this. The question is not asking me to perform the entire proof, just to identify the assumption needed for the proof by contradiction.So, regardless of whether the original statement is true or not, in the context of the problem, I need to assume the opposite to use contradiction.Therefore, the original statement is: "Given ( a ) and ( b ) are real numbers, the equation ( x^2 + ax + b = 0 ) has at least one real root."The negation is: "Given ( a ) and ( b ) are real numbers, the equation ( x^2 + ax + b = 0 ) has no real roots."So, the assumption needed is option A.Therefore, even though in reality, the original statement is not always true, for the sake of the problem, I need to proceed with the assumption that the equation has no real roots, which is option A.So, to sum up, in a proof by contradiction, I assume the opposite of what I want to prove. The original statement claims that the equation has at least one real root. The opposite is that it has no real roots. Therefore, the assumption is option A.

💡Okay, so I have this problem here where I need to find how many ounces of water, y, to add to n ounces of an n% acid solution to get a (n-20)% solution. And it's given that n is greater than 30. Hmm, let me try to break this down step by step.First, I know that when you add water to a solution, you're diluting it, which means the concentration of the acid decreases. In this case, the concentration is decreasing by 20%, so from n% to (n-20)%. That makes sense because adding water doesn't change the amount of acid, only the total volume of the solution.So, let's think about the amount of acid before and after adding the water. Initially, I have n ounces of a solution that's n% acid. To find the amount of acid, I can calculate n% of n ounces. That would be (n/100) * n, which simplifies to n²/100 ounces of acid. Okay, so that's the amount of acid before adding water.After adding y ounces of water, the total volume of the solution becomes n + y ounces. The concentration of the acid is now (n - 20)%, so the amount of acid in this new solution is (n - 20)% of (n + y) ounces. That would be ((n - 20)/100) * (n + y) ounces of acid.Since the amount of acid doesn't change when we add water, these two expressions should be equal. So, I can set up the equation:n²/100 = ((n - 20)/100) * (n + y)Hmm, okay, let me write that out:n²/100 = (n - 20)(n + y)/100Since both sides have a denominator of 100, I can multiply both sides by 100 to eliminate the denominators:n² = (n - 20)(n + y)Alright, now I need to solve for y. Let's expand the right side of the equation:n² = n(n + y) - 20(n + y)n² = n² + ny - 20n - 20yNow, let's subtract n² from both sides to simplify:0 = ny - 20n - 20yHmm, so:ny - 20n - 20y = 0I can factor out terms to solve for y. Let's see, I can factor out n from the first two terms and y from the last term:n(y - 20) - 20y = 0Wait, that doesn't seem quite right. Let me try another approach. Let's collect like terms:ny - 20y = 20nFactor out y from the left side:y(n - 20) = 20nNow, solve for y:y = (20n)/(n - 20)Okay, so that gives me y in terms of n. Let me check if this makes sense. Since n > 30, the denominator n - 20 will be positive, so y will be positive, which makes sense because we're adding water. Also, if n is much larger, say n approaches infinity, y approaches 20n/n, which is 20. So, that seems reasonable.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the equation:n² = (n - 20)(n + y)Expanding the right side:n² = n² + ny - 20n - 20ySubtract n² from both sides:0 = ny - 20n - 20yThen, moving terms around:ny - 20y = 20nFactor out y:y(n - 20) = 20nDivide both sides by (n - 20):y = 20n / (n - 20)Yes, that seems correct. So, the value of y is 20n divided by (n - 20). Looking back at the options given:A) 20n/(n - 20)B) 20n/(n + 20)C) 10n/(n - 10)D) 20n/(n - 30)E) 10n/(n + 10)So, the correct answer should be option A.Wait a second, but I recall that when you add water, the concentration decreases, so the denominator should be larger, right? Because you're adding more volume. Let me think about this again.If I have n ounces of solution and I add y ounces of water, the total volume becomes n + y. The concentration becomes (n - 20)%, so the amount of acid remains the same, which is n²/100. So, setting up the equation again:n²/100 = (n - 20)(n + y)/100Multiplying both sides by 100:n² = (n - 20)(n + y)Expanding:n² = n² + ny - 20n - 20ySubtracting n²:0 = ny - 20n - 20yRearranging:ny - 20y = 20nFactoring:y(n - 20) = 20nDividing:y = 20n / (n - 20)Yes, that still holds. So, I think my initial solution is correct. Maybe I was just overthinking it. So, the answer is A.But wait, let me test it with some numbers to make sure. Let's say n = 40, which is greater than 30. Then, y should be 20*40/(40 - 20) = 800/20 = 40 ounces.So, if I have 40 ounces of a 40% acid solution, that's 16 ounces of acid. Adding 40 ounces of water makes the total volume 80 ounces. The new concentration should be 40 - 20 = 20%. So, 20% of 80 ounces is 16 ounces of acid, which matches. So, that works.Another test: n = 50. Then y = 20*50/(50 - 20) = 1000/30 ≈ 33.33 ounces.Initial acid: 50 ounces of 50% solution is 25 ounces of acid. Adding 33.33 ounces of water makes total volume 83.33 ounces. The new concentration should be 30%. 30% of 83.33 is approximately 25 ounces, which matches. So, that works too.Okay, so my formula seems to be correct. Therefore, the answer is A) 20n/(n - 20).

💡Okay, so I've got this problem about coloring the vertices of a polygon with an odd number of sides, specifically (2n + 1) sides. The condition is that no two neighboring vertices can have the same color. I need to show that it's possible to divide this polygon into triangles using diagonals whose endpoints are of different colors and that these diagonals don't intersect. Hmm, sounds a bit tricky, but let me try to break it down.First, I know that a polygon with (2n + 1) sides is an odd-sided polygon. Since it's colored such that no two adjacent vertices share the same color, that means the coloring must alternate between two colors. Wait, but since it's odd, the number of vertices is odd, so if I start with one color, the last vertex will end up with the same color as the first one. That might cause a problem because they are adjacent, right? But the problem states that two neighboring vertices are never of the same color, so maybe the coloring isn't just a simple two-coloring? Or perhaps it's allowed to have more than two colors? Hmm, the problem doesn't specify the number of colors, just that neighboring vertices are different. Maybe it's possible with two colors because of the way the polygon is structured? I'm not sure, but maybe I don't need to worry about the number of colors right now.Next, the goal is to divide the polygon into triangles using diagonals. I remember that any convex polygon can be triangulated into (n - 2) triangles, where (n) is the number of sides. But in this case, the polygon might not necessarily be convex, but I think the problem doesn't specify, so maybe it's assuming convexity? Or maybe the result holds regardless of convexity. I'll assume convexity for simplicity unless told otherwise.The key here is that the diagonals must have endpoints of different colors and must not intersect. So, I need to find a way to draw these diagonals such that each diagonal connects two vertices of different colors, and none of these diagonals cross each other inside the polygon.Let me think about smaller cases to get an idea. Let's start with (n = 1), so the polygon has (2(1) + 1 = 3) sides, which is a triangle. In this case, there's nothing to triangulate because it's already a triangle. So, that's trivial.What about (n = 2), so a pentagon with 5 sides. Let's color the vertices alternately, say, red and blue. Since it's a pentagon, starting with red, the colors would go red, blue, red, blue, red. Wait, but the first and last vertices are both red and adjacent, which violates the condition. So, maybe it's not possible to color a pentagon with two colors without having two adjacent vertices of the same color? That seems contradictory to the problem statement. Maybe the problem allows more than two colors? Or perhaps I'm misunderstanding the coloring condition.Wait, the problem says "we color all the vertices... such that two neighboring vertices are never of the same color." It doesn't specify the number of colors, so maybe it's using more than two colors. For example, in a pentagon, we can color it with three colors, say red, blue, green, repeating as necessary. So, the coloring would be red, blue, green, red, blue. Then, no two adjacent vertices share the same color. That works. So, maybe the number of colors isn't restricted to two, but just that adjacent vertices are different.Okay, so for a pentagon, I can color it with three colors without any two adjacent vertices sharing the same color. Now, can I triangulate it with diagonals connecting different colors without intersections? Let's see.In a pentagon, triangulation requires two diagonals. Let's pick two non-crossing diagonals. For example, if I connect vertex 1 to 3 and vertex 1 to 4, but wait, vertex 1 is connected to both 3 and 4, but in a pentagon, connecting 1 to 3 and 1 to 4 would create two triangles: 1-2-3 and 1-3-4-5? Wait, no, actually, connecting 1 to 3 and 1 to 4 would create triangles 1-2-3, 1-3-4, and 1-4-5. But that's three triangles, which is more than needed. Wait, no, a pentagon can be triangulated into three triangles with two diagonals. So, maybe connecting 1-3 and 1-4 is correct.But in terms of colors, vertex 1 is, say, red. Vertex 3 is green, and vertex 4 is blue. So, connecting red to green and red to blue. So, both diagonals connect different colors. Also, these diagonals don't cross because they both emanate from vertex 1. So, that works for the pentagon.Okay, so for (n = 2), it's possible. Let's try (n = 3), which is a heptagon with 7 sides. Coloring it with three colors, say red, blue, green, repeating. So, the colors would be red, blue, green, red, blue, green, red. Again, the first and last vertices are red and adjacent, which is a problem. So, maybe I need four colors? Let me try: red, blue, green, yellow, red, blue, green. Then, the last vertex is green, which is different from the first vertex, red. So, that works. So, with four colors, I can color a heptagon without adjacent vertices sharing the same color.Now, can I triangulate this heptagon with diagonals connecting different colors without intersections? Triangulating a heptagon requires five triangles, so four diagonals. I need to draw four non-crossing diagonals connecting vertices of different colors.Let me try to visualize the heptagon. Let's label the vertices 1 through 7, colored as follows: 1 - red, 2 - blue, 3 - green, 4 - yellow, 5 - red, 6 - blue, 7 - green.I need to draw diagonals that connect different colors. Let's start by connecting vertex 1 (red) to vertex 3 (green). That's a diagonal of different colors. Then, from vertex 1, can I connect to another vertex? If I connect 1 to 4 (yellow), that's also different. But connecting 1 to 4 would cross the diagonal 1-3 if I connect 1 to 4. Wait, no, in a heptagon, connecting 1 to 3 and 1 to 4 are both non-crossing because they both start from 1. So, maybe that's okay.But wait, triangulating a heptagon requires four diagonals, so I need to add more. After connecting 1-3 and 1-4, I have split the heptagon into smaller polygons: a triangle 1-2-3, a quadrilateral 1-3-4-5, and a pentagon 1-4-5-6-7. Hmm, but I need to triangulate the entire heptagon, so maybe I need to connect other diagonals as well.Alternatively, maybe I should use a different approach. Perhaps instead of connecting from vertex 1, I can connect other vertices. For example, connect vertex 2 (blue) to vertex 4 (yellow). That's a diagonal of different colors. Then, connect vertex 4 (yellow) to vertex 6 (blue). That's another diagonal of different colors. Then, connect vertex 6 (blue) to vertex 1 (red). That's another diagonal of different colors. Wait, but connecting 6 to 1 would cross the diagonal 2-4 if they are not adjacent. Hmm, maybe not.Alternatively, maybe connect vertex 3 (green) to vertex 5 (red). That's a diagonal of different colors. Then, connect vertex 5 (red) to vertex 7 (green). That's another diagonal of different colors. Then, connect vertex 7 (green) to vertex 2 (blue). That's another diagonal of different colors. But now, I have three diagonals, and I need one more. Maybe connect vertex 2 (blue) to vertex 4 (yellow). But then, does that cross any existing diagonals? Let me see: 2-4, 3-5, 5-7, 7-2. Hmm, 2-4 and 5-7 might cross each other. So, that's a problem.Maybe I need a different strategy. Perhaps start by connecting vertex 1 to vertex 4, which is red to yellow. Then, connect vertex 4 to vertex 7, which is yellow to green. Then, connect vertex 7 to vertex 3, which is green to green? Wait, vertex 3 is green, and vertex 7 is green. That's the same color, so that's not allowed. So, I can't connect 7 to 3.Alternatively, connect vertex 7 to vertex 2, which is green to blue. That's different. So, 7-2 is okay. Then, connect vertex 2 to vertex 5, which is blue to red. That's different. So, now I have diagonals 1-4, 4-7, 7-2, and 2-5. Let me check if these cross. 1-4 and 4-7 don't cross. 4-7 and 7-2 don't cross. 7-2 and 2-5 don't cross. 2-5 and 1-4 might cross? Let me visualize: 1-4 is a diagonal from 1 to 4, and 2-5 is from 2 to 5. In a heptagon, these diagonals would cross each other inside the polygon. So, that's a problem.Hmm, maybe I need to avoid connecting 2-5. Instead, after connecting 1-4, 4-7, and 7-2, maybe connect vertex 2 to vertex 6, which is blue to blue. Wait, same color, not allowed. Alternatively, connect vertex 2 to vertex 5, but that causes crossing. Maybe connect vertex 5 to vertex 3, which is red to green. That's different. So, 5-3. Does that cross any existing diagonals? 5-3 and 1-4: in a heptagon, 1-4 is from 1 to 4, and 5-3 is from 5 to 3. These might cross. Let me see: 1-4 goes from 1 to 4, and 5-3 goes from 5 to 3. Depending on the order, they might cross. So, maybe not.This is getting complicated. Maybe I need a different approach. Perhaps using induction. The problem seems to suggest that for any (2n + 1) polygon, this is possible. So, maybe I can use induction on (n).Base case: (n = 1), which is a triangle. As we saw earlier, it's trivial because it's already a triangle. So, the base case holds.Inductive step: Assume that for a polygon with (2k + 1) sides, it's possible to triangulate it with non-crossing diagonals connecting different colors. Now, consider a polygon with (2(k + 1) + 1 = 2k + 3) sides. I need to show that it's also possible.How can I relate a polygon with (2k + 3) sides to one with (2k + 1) sides? Maybe by finding a diagonal that splits the polygon into a triangle and a polygon with (2k + 1) sides. If I can find such a diagonal with endpoints of different colors, then I can apply the induction hypothesis to the smaller polygon.But how do I ensure that such a diagonal exists? Since the polygon is colored such that no two adjacent vertices have the same color, maybe there's always a vertex whose adjacent vertices are of different colors, allowing me to draw a diagonal to a non-adjacent vertex of a different color.Wait, in a polygon with an odd number of sides, if I color it such that no two adjacent vertices share the same color, then the coloring must alternate between two colors, but since it's odd, the first and last vertices will have the same color, which is adjacent. So, that's a problem. Wait, but the problem states that two neighboring vertices are never of the same color, so maybe the coloring isn't just two colors. Maybe it's more than two colors.Wait, the problem doesn't specify the number of colors, just that adjacent vertices are different. So, maybe it's possible to color it with more than two colors such that no two adjacent vertices share the same color. For example, using three colors for a pentagon, as I did earlier.So, in general, for a polygon with (2n + 1) sides, we can color it with (n + 1) colors such that no two adjacent vertices share the same color. Wait, is that necessary? Or maybe just three colors are sufficient? I'm not sure, but perhaps it's not necessary to know the exact number of colors, just that such a coloring exists.Going back to the inductive step, assuming that for (2k + 1) sides, it's possible, how do I extend it to (2k + 3) sides? Maybe by finding a vertex whose adjacent vertices are of different colors, allowing me to draw a diagonal to a vertex of a different color, splitting the polygon into a triangle and a smaller polygon.Wait, but in a polygon with (2k + 3) sides, if I can find a vertex (v) such that its two neighbors are of different colors, then I can connect (v) to a vertex of a different color, say (w), which is not adjacent to (v), creating a triangle (v)-neighbor1-(w) and a smaller polygon.But I need to ensure that (v) and (w) are of different colors. Since (v) is connected to two neighbors of different colors, maybe (w) can be chosen such that it's of a different color than (v).Alternatively, maybe I can use the fact that the polygon has an odd number of sides, so there must be a vertex whose two neighbors are of different colors, allowing me to draw a diagonal to another vertex of a different color.Wait, let me think about the coloring. Since the polygon has an odd number of sides, and adjacent vertices are colored differently, the coloring must alternate between two colors, but since it's odd, the first and last vertices will have the same color, which is adjacent. So, that's a contradiction unless we use more than two colors.Therefore, the coloring must use at least three colors. So, maybe for a polygon with (2n + 1) sides, we can color it with three colors such that no two adjacent vertices share the same color.Given that, perhaps I can always find a vertex whose two neighbors are of different colors, allowing me to draw a diagonal to another vertex of a different color, splitting the polygon into a triangle and a smaller polygon.So, in the inductive step, I can assume that such a diagonal exists, split the polygon, and then apply the induction hypothesis to the smaller polygon.Therefore, by induction, the statement holds for all (n).Wait, but I feel like I'm hand-waving a bit here. Maybe I need to be more precise.Let me try to formalize the inductive step.Assume that for any polygon with (2k + 1) sides, it's possible to triangulate it with non-crossing diagonals connecting different colors.Now, consider a polygon with (2(k + 1) + 1 = 2k + 3) sides. Since it's colored such that no two adjacent vertices share the same color, and it's an odd polygon, the coloring must use at least three colors.In such a coloring, there must exist at least one vertex whose two neighbors are of different colors. Let's call this vertex (v). Since (v) has two neighbors of different colors, say color A and color B, (v) itself must be of a different color, say color C.Now, from vertex (v), I can draw a diagonal to another vertex (w) such that (w) is of a different color than (v). Since the polygon is colored with at least three colors, there must be a vertex (w) of a different color than (v).By choosing (w) appropriately, I can ensure that the diagonal (v-w) does not cross any other diagonals. Once I draw this diagonal, it splits the polygon into a triangle (v)-neighbor1-(w) and a smaller polygon with (2k + 1) sides.By the induction hypothesis, this smaller polygon can be triangulated with non-crossing diagonals connecting different colors. Therefore, the entire original polygon can be triangulated as required.Thus, by induction, the statement holds for all (n).I think that makes sense. The key idea is that in an odd-sided polygon with a proper coloring, there's always a vertex whose neighbors are of different colors, allowing us to split the polygon into a triangle and a smaller polygon, and then apply the induction hypothesis.I should probably check this with another example, like a heptagon, to make sure it works.Take a heptagon with vertices colored as follows: 1 - red, 2 - blue, 3 - green, 4 - yellow, 5 - red, 6 - blue, 7 - green.Looking for a vertex whose neighbors are of different colors. Let's check vertex 1: neighbors are 7 (green) and 2 (blue). Different colors. So, vertex 1 is red, neighbors are green and blue. So, I can connect vertex 1 to another vertex of a different color, say vertex 4 (yellow). Drawing diagonal 1-4, which connects red to yellow.This splits the heptagon into triangle 1-2-3-4 (wait, no, triangle 1-2-3 and quadrilateral 1-3-4-5-6-7). Wait, no, connecting 1-4 splits the heptagon into triangle 1-2-3-4? No, actually, connecting 1-4 would split it into triangle 1-2-3-4? Wait, no, in a heptagon, connecting 1-4 would create triangle 1-2-3-4? No, actually, connecting 1-4 would create triangle 1-2-3-4? Wait, no, connecting 1-4 would create triangle 1-2-3-4? No, actually, connecting 1-4 would create triangle 1-2-3-4? Wait, I'm getting confused.Wait, in a heptagon, connecting vertex 1 to vertex 4 would create triangle 1-2-3-4? No, actually, it would create triangle 1-2-3 and quadrilateral 1-3-4-5-6-7. Wait, no, connecting 1-4 would create triangle 1-2-3 and a pentagon 1-3-4-5-6-7. Wait, no, a heptagon has seven sides, so connecting 1-4 would split it into a triangle 1-2-3-4? No, that's four vertices, which is a quadrilateral. Wait, no, connecting 1-4 in a heptagon would split it into a triangle 1-2-3-4? No, actually, connecting 1-4 would create triangle 1-2-3 and a pentagon 1-3-4-5-6-7. Wait, that's six vertices, which is a hexagon. Hmm, I'm getting confused.Wait, maybe I need to count the vertices. Original heptagon has vertices 1-7. Connecting 1-4 splits it into two polygons: one is triangle 1-2-3-4? No, triangle has three vertices. So, connecting 1-4 would create triangle 1-2-3 and a quadrilateral 1-3-4-5-6-7? Wait, no, that's six vertices, which is a hexagon. So, actually, connecting 1-4 would split the heptagon into a triangle 1-2-3 and a hexagon 1-3-4-5-6-7.But a hexagon has six sides, which is even. Wait, but our induction hypothesis is for odd-sided polygons. So, maybe I need a different approach.Alternatively, maybe I should connect vertex 1 to vertex 5. Vertex 1 is red, vertex 5 is red. Same color, not allowed. So, can't connect 1-5.Vertex 1 is red, vertex 6 is blue. So, connecting 1-6 would be red to blue, different colors. Let's try that. Connecting 1-6 splits the heptagon into triangle 1-2-3-4-5-6 and a triangle 1-6-7. Wait, no, connecting 1-6 would create triangle 1-2-3-4-5-6? No, that's six vertices, which is a hexagon, and triangle 1-6-7.But again, the hexagon is even-sided, which complicates things because our induction hypothesis is for odd-sided polygons.Hmm, maybe I need to connect vertex 1 to vertex 3. Vertex 1 is red, vertex 3 is green. Different colors. So, connecting 1-3 splits the heptagon into triangle 1-2-3 and a pentagon 1-3-4-5-6-7. Wait, that's six vertices, which is a hexagon. Again, even-sided.Wait, maybe I'm approaching this wrong. Maybe instead of connecting from vertex 1, I should connect from another vertex.Let's look for a vertex whose neighbors are of different colors. Vertex 2 is blue, neighbors are 1 (red) and 3 (green). Different colors. So, vertex 2 is blue, neighbors are red and green. So, I can connect vertex 2 to another vertex of a different color, say vertex 5 (red). Connecting 2-5, which is blue to red, different colors.This splits the heptagon into triangle 2-3-4-5 and a pentagon 1-2-5-6-7. Wait, no, connecting 2-5 would create triangle 2-3-4-5? No, triangle has three vertices. So, connecting 2-5 would create triangle 2-3-4-5? No, that's four vertices, which is a quadrilateral. Wait, no, connecting 2-5 would create triangle 2-3-4 and a pentagon 2-4-5-6-7. Wait, no, connecting 2-5 would create triangle 2-3-4 and a pentagon 2-4-5-6-7. So, the pentagon has five sides, which is odd. Perfect, now I can apply the induction hypothesis to the pentagon.So, in this case, connecting 2-5 splits the heptagon into triangle 2-3-4 and pentagon 2-4-5-6-7. The pentagon can be triangulated by the induction hypothesis, and the triangle is already a triangle. So, overall, the heptagon can be triangulated as required.Therefore, the inductive step holds, and by induction, the statement is true for all (n).I think this makes sense. The key is to find a vertex whose neighbors are of different colors, allowing us to split the polygon into a triangle and a smaller odd-sided polygon, to which we can apply the induction hypothesis.

💡Okay, so I have this math problem here: If x divided by y equals 2, then what is (x minus y) divided by x? The options are (a) -1, (b) -1/2, (c) 1/2, (d) 1, and (e) 2. Hmm, let's see how I can approach this.First, I know that x/y = 2. That means x is twice as big as y. Maybe I can express x in terms of y or vice versa. Let me write that down: x = 2y. Yeah, that makes sense because if I divide x by y, I get 2.Now, the expression I need to find is (x - y)/x. Maybe I can substitute x with 2y in this expression. Let me try that. So, replacing x with 2y, the numerator becomes (2y - y), which simplifies to y. And the denominator is still x, which is 2y. So now, the expression is y/(2y).Wait, y divided by 2y. I can simplify that by canceling out the y in the numerator and denominator, right? So y divided by 2y is 1/2. That seems straightforward.Let me double-check to make sure I didn't make any mistakes. Starting from x/y = 2, so x = 2y. Then, (x - y)/x becomes (2y - y)/2y, which is y/2y, which simplifies to 1/2. Yeah, that looks correct.Alternatively, maybe I can approach it another way to confirm. Let's see, (x - y)/x can be rewritten as x/x - y/x. Since x/x is 1, that simplifies to 1 - y/x. From the original equation, x/y = 2, so y/x would be 1/2. Therefore, 1 - 1/2 is 1/2. Yep, same result.I think I'm confident that the answer is 1/2, which is option (c). It makes sense because if x is twice y, then subtracting y from x leaves y, and dividing that by x (which is 2y) gives half.Just to visualize it, let's pick actual numbers. Suppose y is 1, then x would be 2. So, (x - y)/x becomes (2 - 1)/2, which is 1/2. If y is 2, x is 4, so (4 - 2)/4 is 2/4, which simplifies to 1/2. It consistently gives the same answer regardless of the value of y, as long as x is twice y.I don't see any other way this could go. Maybe if I messed up the substitution or the simplification, but I checked it twice, and both methods gave me the same result. So, I think I'm good.**Final Answer**boxed{C}

💡Okay, so I have this quadratic equation: ( ax^2 + 24x + c = 0 ). The problem says it has exactly one solution. Hmm, I remember that for a quadratic equation to have exactly one solution, the discriminant must be zero. The discriminant formula is ( b^2 - 4ac ). In this case, ( b ) is 24, so let me calculate that.First, let's write down the discriminant:( 24^2 - 4ac = 0 )Calculating ( 24^2 ) gives me 576. So,( 576 - 4ac = 0 )That means ( 4ac = 576 ). Dividing both sides by 4,( ac = 144 )Alright, so the product of ( a ) and ( c ) is 144. The problem also tells me that ( a + c = 31 ). So now I have two equations:1. ( a + c = 31 )2. ( ac = 144 )I need to find the values of ( a ) and ( c ) that satisfy both equations. Since ( a ) and ( c ) are coefficients in a quadratic equation, they can be positive or negative, but given that ( a + c = 31 ) and ( a < c ), I think they are both positive. Let me assume that for now.To solve these equations, I can set up a quadratic equation where ( a ) and ( c ) are the roots. The standard form is:( t^2 - (a + c)t + ac = 0 )Plugging in the known values:( t^2 - 31t + 144 = 0 )Now, I need to factor this quadratic equation. Let me think about two numbers that multiply to 144 and add up to 31. Hmm, 9 and 22 come to mind because 9 * 16 is 144, but 9 + 16 is 25, which is not 31. Wait, 9 * 16 is 144, but that's not helpful. Let me try 9 and 22.9 * 22 is 198, which is too big. Wait, no, that's not right. Wait, 9 * 16 is 144, but 9 + 16 is 25. Hmm, maybe I need to think differently.Wait, 144 divided by 9 is 16, but that doesn't add up to 31. Let me try 12 and 12. 12 * 12 is 144, but 12 + 12 is 24, which is less than 31. Hmm, not helpful.Wait, maybe I made a mistake earlier. Let me try 16 and 9. 16 + 9 is 25, which is still not 31. Hmm, maybe I need to try different factors.Let me list the factor pairs of 144:1 and 144 (sum 145)2 and 72 (sum 74)3 and 48 (sum 51)4 and 36 (sum 40)6 and 24 (sum 30)8 and 18 (sum 26)9 and 16 (sum 25)12 and 12 (sum 24)Wait, none of these add up to 31. Did I miss something? Hmm, maybe I need to consider that ( a ) and ( c ) could be fractions or something else. But since the problem doesn't specify, I think they are integers.Wait, maybe I made a mistake in setting up the quadratic equation. Let me double-check.I have ( a + c = 31 ) and ( ac = 144 ). So the quadratic equation should be ( t^2 - 31t + 144 = 0 ). That seems correct.Let me try to factor it again. Maybe I need to use the quadratic formula to find the roots. The quadratic formula is ( t = frac{-b pm sqrt{b^2 - 4ac}}{2a} ). In this case, ( a = 1 ), ( b = -31 ), and ( c = 144 ).So,( t = frac{31 pm sqrt{(-31)^2 - 4*1*144}}{2*1} )Calculating the discriminant:( (-31)^2 = 961 )( 4*1*144 = 576 )So,( sqrt{961 - 576} = sqrt{385} )Wait, 385 is not a perfect square. That means the roots are not integers. But I was expecting integer solutions because ( a ) and ( c ) are likely integers given the problem statement.Hmm, maybe I made a mistake in the discriminant earlier. Let me go back.The original discriminant was ( 24^2 - 4ac = 0 ), which is ( 576 - 4ac = 0 ), so ( 4ac = 576 ), hence ( ac = 144 ). That seems correct.And ( a + c = 31 ). So, the quadratic equation is correct. Maybe I need to use the quadratic formula despite the roots not being integers.So,( t = frac{31 pm sqrt{385}}{2} )But this would give me two irrational numbers, which seems odd for the problem. Maybe I need to reconsider my approach.Wait, perhaps I made a mistake in the initial setup. Let me check again.The quadratic equation is ( ax^2 + 24x + c = 0 ). The discriminant is ( b^2 - 4ac ), which is ( 24^2 - 4ac = 0 ). So, ( 576 - 4ac = 0 ), leading to ( ac = 144 ). That's correct.Given ( a + c = 31 ), so I have two equations:1. ( a + c = 31 )2. ( ac = 144 )I can solve this system of equations. Let me express ( c ) in terms of ( a ) from the first equation: ( c = 31 - a ). Then substitute into the second equation:( a(31 - a) = 144 )Expanding this,( 31a - a^2 = 144 )Rearranging,( -a^2 + 31a - 144 = 0 )Multiplying both sides by -1,( a^2 - 31a + 144 = 0 )Now, let's try to factor this quadratic equation. Looking for two numbers that multiply to 144 and add up to 31.Wait, earlier I thought 9 and 22 add up to 31, but 9 * 22 is 198, which is not 144. Hmm, that doesn't work.Wait, maybe I need to think differently. Let me try 16 and 9. 16 + 9 = 25, which is not 31. 12 and 12 add up to 24. 18 and 8 add up to 26. 24 and 6 add up to 30. 36 and 4 add up to 40. 48 and 3 add up to 51. 72 and 2 add up to 74. 144 and 1 add up to 145.Wait, none of these add up to 31. So, maybe the quadratic doesn't factor nicely, and I need to use the quadratic formula.So, for the equation ( a^2 - 31a + 144 = 0 ), the quadratic formula gives:( a = frac{31 pm sqrt{(-31)^2 - 4*1*144}}{2*1} )Calculating the discriminant:( 31^2 = 961 )( 4*1*144 = 576 )So,( sqrt{961 - 576} = sqrt{385} )Hmm, 385 is 5*77, which is 5*7*11, so it's not a perfect square. That means the solutions are irrational.But the problem seems to expect integer solutions since it's asking for an ordered pair (a, c) without specifying they have to be integers, but given the context, I think they are integers.Wait, maybe I made a mistake in the discriminant earlier. Let me double-check.Original equation: ( ax^2 + 24x + c = 0 )Discriminant: ( 24^2 - 4ac = 0 )So,( 576 - 4ac = 0 )Hence,( 4ac = 576 )( ac = 144 )That's correct.Given ( a + c = 31 ), so we have:( a + c = 31 )( ac = 144 )I think I need to solve this system. Let me express ( c = 31 - a ) and substitute into ( ac = 144 ):( a(31 - a) = 144 )Which simplifies to:( 31a - a^2 = 144 )Rearranged:( a^2 - 31a + 144 = 0 )This is the same quadratic as before. Since it doesn't factor nicely, I have to use the quadratic formula.So,( a = frac{31 pm sqrt{31^2 - 4*1*144}}{2} )Calculating the discriminant:( 31^2 = 961 )( 4*1*144 = 576 )So,( sqrt{961 - 576} = sqrt{385} )Thus,( a = frac{31 pm sqrt{385}}{2} )Hmm, so ( a ) is either ( frac{31 + sqrt{385}}{2} ) or ( frac{31 - sqrt{385}}{2} ). Since ( a < c ), and ( c = 31 - a ), let's see which one is smaller.If ( a = frac{31 - sqrt{385}}{2} ), then ( c = 31 - a = 31 - frac{31 - sqrt{385}}{2} = frac{62 - 31 + sqrt{385}}{2} = frac{31 + sqrt{385}}{2} ). So, ( a ) is the smaller one, which fits ( a < c ).But these are irrational numbers, which seems odd. Maybe I made a mistake in the problem setup.Wait, let me check the original problem again. It says the quadratic equation ( ax^2 + 24x + c = 0 ) has exactly one solution, ( a + c = 31 ), and ( a < c ). Find the ordered pair ( (a, c) ).Hmm, maybe I need to consider that ( a ) and ( c ) are not necessarily integers. So, perhaps the answer is in terms of radicals.But the problem didn't specify that ( a ) and ( c ) are integers, so maybe that's acceptable. However, the initial thought process assumed integer solutions, which might not be the case.Alternatively, maybe I made a mistake in the discriminant calculation. Let me double-check.Discriminant is ( b^2 - 4ac ). Here, ( b = 24 ), so ( 24^2 = 576 ). Then, ( 4ac = 4*a*c ). Since ( ac = 144 ), ( 4ac = 576 ). So, ( 576 - 576 = 0 ). That's correct.So, the discriminant is zero, which is why there's exactly one solution.Given that, the values of ( a ) and ( c ) are ( frac{31 pm sqrt{385}}{2} ). Since ( a < c ), ( a = frac{31 - sqrt{385}}{2} ) and ( c = frac{31 + sqrt{385}}{2} ).But this seems complicated, and the problem might expect integer solutions. Maybe I need to re-examine my approach.Wait, perhaps I made a mistake in setting up the quadratic equation. Let me try a different method.Given ( a + c = 31 ) and ( ac = 144 ), I can think of ( a ) and ( c ) as the roots of the quadratic equation ( t^2 - 31t + 144 = 0 ). But since this doesn't factor nicely, maybe I need to use completing the square.Let me try that.Starting with ( t^2 - 31t + 144 = 0 )Move the constant term to the other side:( t^2 - 31t = -144 )To complete the square, take half of the coefficient of ( t ), which is ( -31/2 ), square it: ( (31/2)^2 = 961/4 )Add this to both sides:( t^2 - 31t + 961/4 = -144 + 961/4 )Simplify the right side:Convert -144 to fourths: -144 = -576/4So,( -576/4 + 961/4 = (961 - 576)/4 = 385/4 )Now, the left side is a perfect square:( (t - 31/2)^2 = 385/4 )Take square roots of both sides:( t - 31/2 = pm sqrt{385}/2 )So,( t = 31/2 pm sqrt{385}/2 )Which is the same as:( t = frac{31 pm sqrt{385}}{2} )So, the solutions are the same as before. Therefore, ( a ) and ( c ) are ( frac{31 - sqrt{385}}{2} ) and ( frac{31 + sqrt{385}}{2} ), respectively.But this seems complicated, and I suspect the problem expects integer solutions. Maybe I made a mistake in the discriminant calculation.Wait, let me check again. The discriminant is ( b^2 - 4ac ). Here, ( b = 24 ), so ( 24^2 = 576 ). Then, ( 4ac = 4*a*c ). Since ( ac = 144 ), ( 4ac = 576 ). So, ( 576 - 576 = 0 ). That's correct.So, the discriminant is zero, which is why there's exactly one solution.Given that, the values of ( a ) and ( c ) are ( frac{31 pm sqrt{385}}{2} ). Since ( a < c ), ( a = frac{31 - sqrt{385}}{2} ) and ( c = frac{31 + sqrt{385}}{2} ).But this seems complicated, and the problem might expect integer solutions. Maybe I need to re-examine the problem statement.Wait, the problem says "the quadratic equation ( ax^2 + 24x + c = 0 ) has exactly one solution. If ( a + c = 31 ), and ( a < c ), find the ordered pair ( (a, c) )."It doesn't specify that ( a ) and ( c ) are integers, so maybe the answer is indeed in terms of radicals. However, the initial thought process assumed integer solutions, which might not be the case.Alternatively, perhaps I made a mistake in the discriminant calculation. Let me double-check.Discriminant is ( b^2 - 4ac ). Here, ( b = 24 ), so ( 24^2 = 576 ). Then, ( 4ac = 4*a*c ). Since ( ac = 144 ), ( 4ac = 576 ). So, ( 576 - 576 = 0 ). That's correct.So, the discriminant is zero, which is why there's exactly one solution.Given that, the values of ( a ) and ( c ) are ( frac{31 pm sqrt{385}}{2} ). Since ( a < c ), ( a = frac{31 - sqrt{385}}{2} ) and ( c = frac{31 + sqrt{385}}{2} ).But this seems complicated, and I suspect the problem might have a typo or expects a different approach. Alternatively, maybe I need to consider that ( a ) and ( c ) are positive integers, and I need to find such pairs.Wait, let me try to list the factor pairs of 144 again, but this time considering that ( a + c = 31 ). Maybe I missed something.Factor pairs of 144:1 and 144 (sum 145)2 and 72 (sum 74)3 and 48 (sum 51)4 and 36 (sum 40)6 and 24 (sum 30)8 and 18 (sum 26)9 and 16 (sum 25)12 and 12 (sum 24)Hmm, none of these add up to 31. So, there are no integer solutions for ( a ) and ( c ) that satisfy both ( a + c = 31 ) and ( ac = 144 ). Therefore, the solutions must be irrational.So, the ordered pair ( (a, c) ) is ( left( frac{31 - sqrt{385}}{2}, frac{31 + sqrt{385}}{2} right) ).But this seems complicated, and I wonder if there's a simpler way to express it or if I made a mistake somewhere.Wait, maybe I can simplify ( sqrt{385} ). Let's see, 385 factors into 5 * 77, which is 5 * 7 * 11. None of these are perfect squares, so ( sqrt{385} ) cannot be simplified further.Therefore, the ordered pair is ( left( frac{31 - sqrt{385}}{2}, frac{31 + sqrt{385}}{2} right) ).But the problem didn't specify that ( a ) and ( c ) are integers, so this should be the correct answer.Alternatively, maybe I need to present it in a different form. Let me calculate the approximate values to check if they make sense.Calculating ( sqrt{385} ):( 19^2 = 361 )( 20^2 = 400 )So, ( sqrt{385} ) is between 19 and 20. Let's approximate it:19.6^2 = 384.1619.6^2 = 384.16, which is very close to 385. So, ( sqrt{385} approx 19.6214 )Therefore,( a = frac{31 - 19.6214}{2} = frac{11.3786}{2} approx 5.6893 )( c = frac{31 + 19.6214}{2} = frac{50.6214}{2} approx 25.3107 )So, ( a approx 5.69 ) and ( c approx 25.31 ). Checking ( a + c approx 5.69 + 25.31 = 31 ), which matches. And ( a * c approx 5.69 * 25.31 approx 144 ), which also matches.Therefore, the ordered pair is ( left( frac{31 - sqrt{385}}{2}, frac{31 + sqrt{385}}{2} right) ).But the problem might expect an exact form rather than decimal approximations. So, I think this is the correct answer.However, I'm a bit confused because the initial thought process assumed integer solutions, but it seems there are no integer solutions. Therefore, the answer must be in terms of radicals.So, to summarize, the ordered pair ( (a, c) ) is ( left( frac{31 - sqrt{385}}{2}, frac{31 + sqrt{385}}{2} right) ).

💡Okay, so I have these two complex number problems to solve. Let me take them one at a time. Starting with the first problem: 1. Given a non-zero complex number ( z ) that satisfies ( |z + 2| = 2 ) and ( z + frac{4}{z} in mathbb{R} ), find the complex number ( z ).Alright, so ( z ) is a complex number, which I can write as ( z = a + bi ), where ( a ) and ( b ) are real numbers, and ( i ) is the imaginary unit. First, the condition ( |z + 2| = 2 ). The modulus of a complex number ( |z| ) is the distance from the origin to the point ( z ) in the complex plane. So, ( |z + 2| = 2 ) means that the distance from ( z ) to ( -2 ) on the real axis is 2. Geometrically, this is a circle centered at ( (-2, 0) ) with radius 2.So, if ( z = a + bi ), then ( |z + 2| = |(a + 2) + bi| = sqrt{(a + 2)^2 + b^2} = 2 ). Squaring both sides, we get ( (a + 2)^2 + b^2 = 4 ). That's equation (1).Next, the condition ( z + frac{4}{z} ) is a real number. Let's compute ( z + frac{4}{z} ).First, ( frac{4}{z} ) can be written as ( frac{4}{a + bi} ). To simplify this, multiply numerator and denominator by the complex conjugate of the denominator:( frac{4}{a + bi} = frac{4(a - bi)}{(a + bi)(a - bi)} = frac{4a - 4bi}{a^2 + b^2} ).So, ( z + frac{4}{z} = (a + bi) + left( frac{4a}{a^2 + b^2} - frac{4b}{a^2 + b^2}i right) ).Combine like terms:Real part: ( a + frac{4a}{a^2 + b^2} ).Imaginary part: ( b - frac{4b}{a^2 + b^2} ).Since ( z + frac{4}{z} ) is real, the imaginary part must be zero. So:( b - frac{4b}{a^2 + b^2} = 0 ).Factor out ( b ):( b left( 1 - frac{4}{a^2 + b^2} right) = 0 ).So, either ( b = 0 ) or ( 1 - frac{4}{a^2 + b^2} = 0 ).Case 1: ( b = 0 ).If ( b = 0 ), then ( z = a ), a real number. Then, from equation (1):( (a + 2)^2 + 0 = 4 ) => ( (a + 2)^2 = 4 ) => ( a + 2 = pm 2 ) => ( a = 0 ) or ( a = -4 ).But the problem states that ( z ) is non-zero. If ( a = 0 ), then ( z = 0 ), which is not allowed. If ( a = -4 ), then ( z = -4 ). Let's check if this satisfies the second condition.Compute ( z + frac{4}{z} ):( -4 + frac{4}{-4} = -4 - 1 = -5 ), which is real. So, ( z = -4 ) is a solution.Wait, but earlier, when I considered ( b = 0 ), I thought ( a = 0 ) is the only solution, but actually, ( a = -4 ) is also a solution. So, ( z = -4 ) is a possible solution.But let's check if ( z = -4 ) satisfies ( |z + 2| = 2 ):( |-4 + 2| = |-2| = 2 ), which is correct. So, ( z = -4 ) is a valid solution.But wait, the problem says "a non-zero complex number", so ( z = -4 ) is non-zero, so that's fine.Case 2: ( 1 - frac{4}{a^2 + b^2} = 0 ).This implies ( a^2 + b^2 = 4 ). So, the modulus of ( z ) is 2.So, ( |z| = 2 ).Now, from equation (1):( (a + 2)^2 + b^2 = 4 ).But since ( a^2 + b^2 = 4 ), substitute into equation (1):( (a + 2)^2 + (4 - a^2) = 4 ).Expand ( (a + 2)^2 ):( a^2 + 4a + 4 + 4 - a^2 = 4 ).Simplify:( 4a + 8 = 4 ) => ( 4a = -4 ) => ( a = -1 ).So, ( a = -1 ). Then, from ( a^2 + b^2 = 4 ):( (-1)^2 + b^2 = 4 ) => ( 1 + b^2 = 4 ) => ( b^2 = 3 ) => ( b = pm sqrt{3} ).Therefore, in this case, ( z = -1 pm sqrt{3}i ).So, altogether, the solutions are ( z = -4 ) and ( z = -1 pm sqrt{3}i ).But wait, the problem says "a non-zero complex number". So, ( z = -4 ) is a real number, but it's also a complex number. So, both ( z = -4 ) and ( z = -1 pm sqrt{3}i ) are solutions.But let me check if ( z = -4 ) satisfies both conditions.First condition: ( |z + 2| = |-4 + 2| = |-2| = 2 ). Yes.Second condition: ( z + frac{4}{z} = -4 + frac{4}{-4} = -4 -1 = -5 ), which is real. So, yes, it works.So, there are three solutions: ( z = -4 ), ( z = -1 + sqrt{3}i ), and ( z = -1 - sqrt{3}i ).But wait, in the initial problem statement, it says "find the complex number ( z )", which might imply a single answer, but since there are multiple solutions, I should list all of them.Wait, but let me think again. When I considered ( b = 0 ), I got ( z = -4 ). When ( a^2 + b^2 = 4 ), I got ( z = -1 pm sqrt{3}i ). So, all three are valid.But let me check if ( z = -4 ) is the only real solution. Yes, because when ( b = 0 ), ( a = -4 ) is the only non-zero solution.So, the solutions are ( z = -4 ), ( z = -1 + sqrt{3}i ), and ( z = -1 - sqrt{3}i ).Wait, but in the problem statement, it says "a non-zero complex number". So, all three are non-zero, so they are all valid.But in the initial thought process, I considered both cases, so I think that's correct.Now, moving on to the second problem:2. Given an imaginary number ( z ) that makes both ( frac{z^2}{z + 1} ) and ( frac{z}{z^2 + 1} ) real numbers, find the imaginary number ( z ).An imaginary number is a complex number with no real part, so ( z = bi ), where ( b ) is a real number, and ( b neq 0 ) (since it's an imaginary number, not zero).So, let me write ( z = bi ).First, compute ( frac{z^2}{z + 1} ).Compute ( z^2 = (bi)^2 = -b^2 ).Compute ( z + 1 = bi + 1 = 1 + bi ).So, ( frac{z^2}{z + 1} = frac{-b^2}{1 + bi} ).To make this real, the imaginary part must be zero.Similarly, compute ( frac{z}{z^2 + 1} ).Compute ( z^2 + 1 = -b^2 + 1 ).So, ( frac{z}{z^2 + 1} = frac{bi}{1 - b^2} ).For this to be real, since the denominator is real (because ( b ) is real), the numerator must be real. But the numerator is ( bi ), which is purely imaginary. So, for ( frac{bi}{1 - b^2} ) to be real, the denominator must be zero, but that would make it undefined. Alternatively, the numerator must be zero, but ( b neq 0 ) because ( z ) is an imaginary number. Wait, that seems contradictory.Wait, perhaps I made a mistake. Let me re-examine.Given ( z = bi ), compute ( frac{z}{z^2 + 1} ).( z^2 = (bi)^2 = -b^2 ).So, ( z^2 + 1 = -b^2 + 1 = 1 - b^2 ).Therefore, ( frac{z}{z^2 + 1} = frac{bi}{1 - b^2} ).For this to be a real number, the imaginary part must be zero. But ( frac{bi}{1 - b^2} ) is purely imaginary unless the denominator is infinite, which is not possible. So, the only way for this to be real is if the numerator is zero, but ( b neq 0 ). Therefore, there is no solution unless the denominator is zero, but that would make it undefined.Wait, that can't be right. Maybe I made a mistake in interpreting the problem.Wait, the problem says "an imaginary number ( z )", which is a complex number with zero real part, so ( z = bi ). So, ( z ) is purely imaginary.But let's see:First, ( frac{z^2}{z + 1} ) must be real.Compute ( z^2 = (bi)^2 = -b^2 ).Compute ( z + 1 = 1 + bi ).So, ( frac{z^2}{z + 1} = frac{-b^2}{1 + bi} ).Multiply numerator and denominator by the conjugate of the denominator:( frac{-b^2 (1 - bi)}{(1 + bi)(1 - bi)} = frac{-b^2 + b^3 i}{1 + b^2} ).So, the expression is ( frac{-b^2}{1 + b^2} + frac{b^3}{1 + b^2}i ).For this to be real, the imaginary part must be zero:( frac{b^3}{1 + b^2} = 0 ).This implies ( b^3 = 0 ), so ( b = 0 ). But ( z ) is an imaginary number, so ( b neq 0 ). Contradiction.Wait, that suggests there is no solution, but the problem says "find the imaginary number ( z )", implying there is a solution. So, perhaps I made a mistake.Wait, let me check the second condition again.Compute ( frac{z}{z^2 + 1} ).( z = bi ), so ( z^2 = -b^2 ), so ( z^2 + 1 = 1 - b^2 ).Thus, ( frac{z}{z^2 + 1} = frac{bi}{1 - b^2} ).For this to be real, the imaginary part must be zero, but ( frac{bi}{1 - b^2} ) is purely imaginary unless ( b = 0 ), which is not allowed. So, this seems impossible.But the problem states that both expressions are real, so perhaps I made a mistake in interpreting ( z ) as purely imaginary. Wait, the problem says "an imaginary number ( z )", which usually means ( z ) has no real part, i.e., ( z = bi ). But maybe it's a complex number with a non-zero imaginary part, but possibly a non-zero real part as well? Wait, no, "imaginary number" typically refers to a complex number with zero real part.Wait, perhaps the problem is using "imaginary number" in a different way. Let me check the problem again."Given an imaginary number ( z ) that makes both ( frac{z^2}{z + 1} ) and ( frac{z}{z^2 + 1} ) real numbers, find the imaginary number ( z )."Hmm, perhaps "imaginary number" here means a complex number with a non-zero imaginary part, not necessarily zero real part. That is, ( z ) is a complex number where the imaginary part is non-zero, but the real part can be non-zero as well. So, ( z = a + bi ) with ( b neq 0 ).If that's the case, then I should let ( z = a + bi ) with ( b neq 0 ).So, let me proceed with that assumption.Let ( z = a + bi ), ( b neq 0 ).First, compute ( frac{z^2}{z + 1} ).Compute ( z^2 = (a + bi)^2 = a^2 - b^2 + 2abi ).Compute ( z + 1 = (a + 1) + bi ).So, ( frac{z^2}{z + 1} = frac{a^2 - b^2 + 2abi}{(a + 1) + bi} ).To make this real, the imaginary part must be zero.Similarly, compute ( frac{z}{z^2 + 1} ).Compute ( z^2 + 1 = (a^2 - b^2 + 1) + 2abi ).So, ( frac{z}{z^2 + 1} = frac{a + bi}{(a^2 - b^2 + 1) + 2abi} ).Again, for this to be real, the imaginary part must be zero.So, I have two conditions:1. The imaginary part of ( frac{z^2}{z + 1} ) is zero.2. The imaginary part of ( frac{z}{z^2 + 1} ) is zero.Let me compute these.First, let me compute ( frac{z^2}{z + 1} ).Let me denote ( z^2 = (a^2 - b^2) + 2abi ), and ( z + 1 = (a + 1) + bi ).So, ( frac{z^2}{z + 1} = frac{(a^2 - b^2) + 2abi}{(a + 1) + bi} ).Multiply numerator and denominator by the conjugate of the denominator:( frac{[(a^2 - b^2) + 2abi][(a + 1) - bi]}{(a + 1)^2 + b^2} ).Let me compute the numerator:First, expand the numerator:( (a^2 - b^2)(a + 1) - (a^2 - b^2)bi + 2abi(a + 1) - 2abi cdot bi ).Simplify term by term:1. ( (a^2 - b^2)(a + 1) = a^3 + a^2 - a b^2 - b^2 ).2. ( - (a^2 - b^2)bi = -a^2 b i + b^3 i ).3. ( 2abi(a + 1) = 2a^2 b i + 2ab i ).4. ( -2abi cdot bi = -2ab i^2 b = -2ab(-1)b = 2ab^2 ).So, combining all terms:Real parts:( a^3 + a^2 - a b^2 - b^2 + 2ab^2 ).Imaginary parts:( (-a^2 b + b^3 + 2a^2 b + 2ab) i ).Simplify real parts:( a^3 + a^2 - a b^2 - b^2 + 2ab^2 = a^3 + a^2 + ab^2 - b^2 ).Imaginary parts:( (-a^2 b + b^3 + 2a^2 b + 2ab) i = (a^2 b + b^3 + 2ab) i ).So, the numerator is ( (a^3 + a^2 + ab^2 - b^2) + (a^2 b + b^3 + 2ab) i ).The denominator is ( (a + 1)^2 + b^2 = a^2 + 2a + 1 + b^2 ).So, ( frac{z^2}{z + 1} = frac{(a^3 + a^2 + ab^2 - b^2) + (a^2 b + b^3 + 2ab) i}{a^2 + 2a + 1 + b^2} ).For this to be real, the imaginary part must be zero:( a^2 b + b^3 + 2ab = 0 ).Factor out ( b ):( b(a^2 + b^2 + 2a) = 0 ).Since ( b neq 0 ), we have:( a^2 + b^2 + 2a = 0 ). Let's call this equation (2).Now, let's compute the second condition: ( frac{z}{z^2 + 1} ) is real.Compute ( z^2 + 1 = (a^2 - b^2 + 1) + 2abi ).So, ( frac{z}{z^2 + 1} = frac{a + bi}{(a^2 - b^2 + 1) + 2abi} ).Multiply numerator and denominator by the conjugate of the denominator:( frac{(a + bi)[(a^2 - b^2 + 1) - 2abi]}{(a^2 - b^2 + 1)^2 + (2ab)^2} ).Compute the numerator:Expand:( a(a^2 - b^2 + 1) - 2a^2 b i + bi(a^2 - b^2 + 1) - 2ab i^2 b ).Simplify term by term:1. ( a(a^2 - b^2 + 1) = a^3 - a b^2 + a ).2. ( -2a^2 b i ).3. ( bi(a^2 - b^2 + 1) = a^2 b i - b^3 i + b i ).4. ( -2ab i^2 b = -2ab(-1)b = 2ab^2 ).Combine all terms:Real parts:( a^3 - a b^2 + a + 2ab^2 ).Imaginary parts:( (-2a^2 b + a^2 b - b^3 + b) i ).Simplify real parts:( a^3 - a b^2 + a + 2ab^2 = a^3 + a + ab^2 ).Imaginary parts:( (-a^2 b - b^3 + b) i ).So, the numerator is ( (a^3 + a + ab^2) + (-a^2 b - b^3 + b) i ).The denominator is ( (a^2 - b^2 + 1)^2 + (2ab)^2 ).For ( frac{z}{z^2 + 1} ) to be real, the imaginary part must be zero:( -a^2 b - b^3 + b = 0 ).Factor out ( b ):( b(-a^2 - b^2 + 1) = 0 ).Since ( b neq 0 ), we have:( -a^2 - b^2 + 1 = 0 ) => ( a^2 + b^2 = 1 ). Let's call this equation (3).Now, we have two equations:From equation (2): ( a^2 + b^2 + 2a = 0 ).From equation (3): ( a^2 + b^2 = 1 ).Subtract equation (3) from equation (2):( (a^2 + b^2 + 2a) - (a^2 + b^2) = 0 - 1 ).Simplify:( 2a = -1 ) => ( a = -frac{1}{2} ).Now, substitute ( a = -frac{1}{2} ) into equation (3):( (-frac{1}{2})^2 + b^2 = 1 ) => ( frac{1}{4} + b^2 = 1 ) => ( b^2 = frac{3}{4} ) => ( b = pm frac{sqrt{3}}{2} ).Therefore, ( z = a + bi = -frac{1}{2} pm frac{sqrt{3}}{2}i ).So, the solutions are ( z = -frac{1}{2} + frac{sqrt{3}}{2}i ) and ( z = -frac{1}{2} - frac{sqrt{3}}{2}i ).Let me verify these solutions.First, check ( frac{z^2}{z + 1} ).Take ( z = -frac{1}{2} + frac{sqrt{3}}{2}i ).Compute ( z + 1 = frac{1}{2} + frac{sqrt{3}}{2}i ).Compute ( z^2 = left(-frac{1}{2}right)^2 - left(frac{sqrt{3}}{2}right)^2 + 2 cdot left(-frac{1}{2}right) cdot frac{sqrt{3}}{2}i = frac{1}{4} - frac{3}{4} - frac{sqrt{3}}{2}i = -frac{1}{2} - frac{sqrt{3}}{2}i ).So, ( frac{z^2}{z + 1} = frac{-frac{1}{2} - frac{sqrt{3}}{2}i}{frac{1}{2} + frac{sqrt{3}}{2}i} ).Multiply numerator and denominator by the conjugate of the denominator:( frac{(-frac{1}{2} - frac{sqrt{3}}{2}i)(frac{1}{2} - frac{sqrt{3}}{2}i)}{(frac{1}{2})^2 + (frac{sqrt{3}}{2})^2} ).Compute numerator:( (-frac{1}{2})(frac{1}{2}) + (-frac{1}{2})(-frac{sqrt{3}}{2}i) + (-frac{sqrt{3}}{2}i)(frac{1}{2}) + (-frac{sqrt{3}}{2}i)(-frac{sqrt{3}}{2}i) ).Simplify:( -frac{1}{4} + frac{sqrt{3}}{4}i - frac{sqrt{3}}{4}i + frac{3}{4}i^2 ).Since ( i^2 = -1 ):( -frac{1}{4} + 0 + frac{3}{4}(-1) = -frac{1}{4} - frac{3}{4} = -1 ).Denominator:( frac{1}{4} + frac{3}{4} = 1 ).So, ( frac{z^2}{z + 1} = -1 ), which is real.Now, check ( frac{z}{z^2 + 1} ).Compute ( z^2 + 1 = -frac{1}{2} - frac{sqrt{3}}{2}i + 1 = frac{1}{2} - frac{sqrt{3}}{2}i ).So, ( frac{z}{z^2 + 1} = frac{-frac{1}{2} + frac{sqrt{3}}{2}i}{frac{1}{2} - frac{sqrt{3}}{2}i} ).Multiply numerator and denominator by the conjugate of the denominator:( frac{(-frac{1}{2} + frac{sqrt{3}}{2}i)(frac{1}{2} + frac{sqrt{3}}{2}i)}{(frac{1}{2})^2 + (frac{sqrt{3}}{2})^2} ).Compute numerator:( (-frac{1}{2})(frac{1}{2}) + (-frac{1}{2})(frac{sqrt{3}}{2}i) + (frac{sqrt{3}}{2}i)(frac{1}{2}) + (frac{sqrt{3}}{2}i)(frac{sqrt{3}}{2}i) ).Simplify:( -frac{1}{4} - frac{sqrt{3}}{4}i + frac{sqrt{3}}{4}i + frac{3}{4}i^2 ).Again, ( i^2 = -1 ):( -frac{1}{4} + 0 + frac{3}{4}(-1) = -frac{1}{4} - frac{3}{4} = -1 ).Denominator is 1 as before.So, ( frac{z}{z^2 + 1} = -1 ), which is real.Similarly, for ( z = -frac{1}{2} - frac{sqrt{3}}{2}i ), the same steps would show that both expressions are real.Therefore, the solutions are ( z = -frac{1}{2} pm frac{sqrt{3}}{2}i ).So, summarizing both problems:1. The solutions are ( z = -4 ), ( z = -1 + sqrt{3}i ), and ( z = -1 - sqrt{3}i ).2. The solutions are ( z = -frac{1}{2} + frac{sqrt{3}}{2}i ) and ( z = -frac{1}{2} - frac{sqrt{3}}{2}i ).But wait, in the first problem, I initially thought ( z = -4 ) was a solution, but when I considered the case ( a^2 + b^2 = 4 ), I got ( z = -1 pm sqrt{3}i ). But when I checked ( z = -4 ), it also satisfied both conditions. So, both are valid.However, in the problem statement, it says "a non-zero complex number", so all three are valid. But perhaps the problem expects all solutions, so I should list all of them.But in the initial thought process, I considered both cases, so I think that's correct.For the second problem, the solutions are ( z = -frac{1}{2} pm frac{sqrt{3}}{2}i ).So, final answers:1. ( z = -4 ), ( z = -1 + sqrt{3}i ), ( z = -1 - sqrt{3}i ).2. ( z = -frac{1}{2} + frac{sqrt{3}}{2}i ), ( z = -frac{1}{2} - frac{sqrt{3}}{2}i ).But wait, in the first problem, when I considered ( b = 0 ), I got ( z = -4 ), but when I considered ( a^2 + b^2 = 4 ), I got ( z = -1 pm sqrt{3}i ). So, all three are valid.But let me check if ( z = -4 ) is indeed a solution.Compute ( |z + 2| = |-4 + 2| = |-2| = 2 ). Yes.Compute ( z + frac{4}{z} = -4 + frac{4}{-4} = -4 -1 = -5 ), which is real. So, yes, it works.Therefore, the solutions for the first problem are ( z = -4 ), ( z = -1 + sqrt{3}i ), and ( z = -1 - sqrt{3}i ).But perhaps the problem expects all solutions, so I should list all of them.In conclusion, the solutions are as above.