💡First, I need to determine the rainfall during the first week. I know the total rainfall for the two weeks is 25 inches, and the rainfall during the second week is 15 inches.By subtracting the second week's rainfall from the total, I can find the first week's rainfall:25 inches (total) - 15 inches (second week) = 10 inches (first week).Now that I have both weeks' rainfall amounts, I can calculate the ratio of the second week's rainfall to the first week's rainfall. This is done by dividing the second week's rainfall by the first week's rainfall:15 inches / 10 inches = 1.5.To express this as a ratio, I can write it as 3:2, since 1.5 is equivalent to 3/2.

💡Okay, so I have this problem where I need to prove that there's a function ( f ) from the set of all natural numbers to itself such that for any natural number ( n ), ( f(f(n)) = n^2 ). Hmm, that sounds interesting. I need to figure out how to define such a function ( f ). First, let me understand what the problem is asking. I need a function that, when applied twice, gives me the square of the original number. So, if I take any natural number ( n ), apply ( f ) to it, and then apply ( f ) again to the result, I should get ( n^2 ). I guess one way to approach this is to think about how functions can be composed. Since applying ( f ) twice gives me ( n^2 ), maybe I can define ( f ) in such a way that it somehow relates to squaring, but not directly. Maybe I can partition the natural numbers into some kind of sequences or cycles where applying ( f ) moves me along the sequence, and applying it twice gets me to the square.Let me try to think of an example. Suppose I start with the number 2. If I apply ( f ) to 2, I get some number, say 3. Then, applying ( f ) to 3 should give me ( 2^2 = 4 ). So, ( f(2) = 3 ) and ( f(3) = 4 ). Now, applying ( f ) to 4 should give me ( 3^2 = 9 ), so ( f(4) = 9 ). Then, ( f(9) ) should be ( 4^2 = 16 ), and so on. This seems to create a chain: 2 → 3 → 4 → 9 → 16 → 81 → ... Okay, so this chain works for 2, 3, 4, 9, 16, etc. But what about other numbers? For example, what happens to 5? If I follow the same pattern, I need to define ( f(5) ) such that ( f(f(5)) = 25 ). Maybe I can pair 5 with another number, say 6. So, ( f(5) = 6 ) and ( f(6) = 25 ). Then, ( f(25) = 36 ) because ( f(f(6)) = 6^2 = 36 ), and ( f(36) = 625 ) since ( f(f(25)) = 25^2 = 625 ), and so on. This seems to work for 5 and 6 as well. So, it looks like I can create pairs of numbers where each pair starts a new chain. For each pair ( (p, q) ), I define ( f(p) = q ) and ( f(q) = p^2 ). Then, ( f(f(p)) = f(q) = p^2 ), which satisfies the condition. Similarly, ( f(f(q)) = f(p^2) ). Wait, I need to make sure that ( f(p^2) ) is defined properly. If I follow the same pattern, ( f(p^2) ) should be ( q^2 ), because ( f(f(q)) = q^2 ). So, ( f(p^2) = q^2 ). Then, ( f(q^2) ) should be ( (p^2)^2 = p^4 ), and so on. This creates an infinite chain for each starting pair ( (p, q) ). But how do I ensure that every natural number is included in some chain? I need to make sure that the function ( f ) is defined for all natural numbers, not just some of them. So, I need to partition the natural numbers into these chains in such a way that every number is part of exactly one chain. One way to do this is to start with the smallest unused number and pair it with the next smallest unused number, then define the chains accordingly. For example, start with 2 and 3, then 5 and 6, then 7 and 8, and so on. This way, every natural number is eventually included in a chain, and the function ( f ) is well-defined for all natural numbers. Let me test this idea with a few more numbers to see if it holds. Take 7 and 8. Define ( f(7) = 8 ) and ( f(8) = 49 ) (since ( 7^2 = 49 )). Then, ( f(49) = 64 ) (since ( 8^2 = 64 )), and ( f(64) = 49^2 = 2401 ), and so on. This seems to work. What about 1? Hmm, 1 is a special case because ( 1^2 = 1 ). So, if I apply ( f ) twice to 1, I should get 1. That means ( f(f(1)) = 1 ). So, ( f(1) ) must be a number that maps back to 1 when ( f ) is applied again. Let's say ( f(1) = 1 ). Then, ( f(f(1)) = f(1) = 1 ), which satisfies the condition. Alternatively, I could pair 1 with another number, but since 1 squared is itself, it's simpler to have ( f(1) = 1 ). So, putting it all together, the function ( f ) can be defined as follows:- For ( n = 1 ), ( f(1) = 1 ).- For each pair of consecutive unused natural numbers ( p ) and ( q ), define ( f(p) = q ) and ( f(q) = p^2 ). Then, continue the chain by defining ( f(p^2) = q^2 ), ( f(q^2) = p^4 ), and so on.This way, every natural number is included in exactly one chain, and applying ( f ) twice to any number ( n ) gives ( n^2 ). I think this construction works, but I need to make sure there are no overlaps or gaps. Since I'm pairing numbers sequentially and defining the chains accordingly, every number should be covered without repetition. Also, since the chains are infinite, there's no issue with running out of numbers. Another thing to consider is whether this function is well-defined. For each ( n ), there's a unique ( f(n) ) based on its position in the chain. There's no ambiguity because each number belongs to only one chain. In summary, by constructing chains of natural numbers where each chain starts with a pair ( (p, q) ) and follows the rule ( f(p) = q ), ( f(q) = p^2 ), ( f(p^2) = q^2 ), etc., we can define a function ( f ) such that ( f(f(n)) = n^2 ) for all natural numbers ( n ). This ensures that the function ( f ) satisfies the given condition for all natural numbers.

💡Okay, so I have this problem here: I need to find the value of 'a' such that the number 2 times 10 to the power of 10 plus 'a' is divisible by 11. The condition is that 'a' is a real number and it's between 0 and 11, including 0 but not reaching 11. Hmm, okay, let me think about how to approach this.First, I remember that there's a rule for checking if a number is divisible by 11. It involves alternating sums of the digits. But wait, in this case, the number isn't given directly; it's given in terms of 10^10, which is a pretty big number. Maybe I can use modular arithmetic instead? I think that might be a better approach here.So, if 2 × 10^10 + a is divisible by 11, that means when we divide this number by 11, the remainder should be zero. In modular terms, that would mean:2 × 10^10 + a ≡ 0 mod 11Which implies:a ≡ -2 × 10^10 mod 11But since we're dealing with mod 11, I can simplify 10^10 mod 11 first. Maybe that will make things easier.I recall that 10 mod 11 is just 10, right? So, 10^1 mod 11 is 10. Then, 10^2 mod 11 would be (10 × 10) mod 11, which is 100 mod 11. Let me calculate that: 11 × 9 = 99, so 100 - 99 = 1. So, 10^2 mod 11 is 1.Wait, that's interesting. So, 10^2 ≡ 1 mod 11. That must mean that 10^even power ≡ 1 mod 11, and 10^odd power ≡ 10 mod 11. Let me test that with 10^3: 10^3 = 1000, and 1000 mod 11. Let's see, 11 × 90 = 990, so 1000 - 990 = 10. Yep, so 10^3 ≡ 10 mod 11. Similarly, 10^4 would be 10^2 × 10^2, which is 1 × 1 = 1 mod 11. So, the pattern repeats every two exponents.Given that, 10^10 is an even exponent, so 10^10 ≡ 1 mod 11. Therefore, 2 × 10^10 ≡ 2 × 1 = 2 mod 11.So, going back to the original equation:2 × 10^10 + a ≡ 0 mod 11Which simplifies to:2 + a ≡ 0 mod 11So, a ≡ -2 mod 11But since 'a' is between 0 and 11, I need to find a positive equivalent of -2 mod 11. To do that, I can add 11 to -2:-2 + 11 = 9So, a ≡ 9 mod 11. Since 9 is within the range [0, 11), that must be the value of 'a'.Wait, let me double-check my steps to make sure I didn't make a mistake. I used the fact that 10^2 ≡ 1 mod 11, so 10^10 is (10^2)^5, which is 1^5 = 1 mod 11. Then, multiplying by 2 gives 2 mod 11. So, 2 + a ≡ 0 mod 11 implies a ≡ -2 mod 11, which is 9. Yeah, that seems correct.Alternatively, I could have considered the divisibility rule for 11, which involves alternating sums of digits. Let's see if that approach would work here as well, just to confirm.The number is 2 × 10^10 + a. Let me write out what 2 × 10^10 looks like. 10^10 is 10000000000, so 2 × 10^10 is 20000000000. Adding 'a' to that would give 20000000000 + a. Since 'a' is less than 11, it's a single digit number or a number with one digit. So, the number would be 2000000000a, where 'a' is the units digit.Now, applying the divisibility rule for 11: subtract the sum of the digits in the odd positions from the sum of the digits in the even positions. If the result is a multiple of 11 (including zero), then the number is divisible by 11.Let's write out the number: 2 0 0 0 0 0 0 0 0 0 aPositions (from right, starting at 1): a is position 1, then the next digit is position 2, and so on up to the first digit which is position 11.Wait, actually, positions are usually counted from the right, starting at 1. So, in the number 2000000000a, the digits are:Position 11: 2Position 10: 0Position 9: 0...Position 2: 0Position 1: aSo, the sum of digits in odd positions: positions 1, 3, 5, 7, 9, 11.Which are: a, 0, 0, 0, 0, 2. So, sum = a + 0 + 0 + 0 + 0 + 2 = a + 2.Sum of digits in even positions: positions 2, 4, 6, 8, 10.Which are: 0, 0, 0, 0, 0. So, sum = 0.Now, the difference between these sums is (a + 2) - 0 = a + 2.For the number to be divisible by 11, this difference must be a multiple of 11, including zero. So, a + 2 must be equal to 0, 11, -11, 22, etc. But since a is between 0 and 11, the only possible value is a + 2 = 11, which gives a = 9.So, both methods give me the same result: a = 9. That makes me more confident that the answer is correct.I think I've covered all the steps and checked my work using two different methods. I don't see any mistakes in my reasoning, so I'm pretty sure that 'a' is 9.**Final Answer**The value of ( a ) is boxed{9}.

💡Okay, so I have this quadratic equation: ( x^2 + 2(a - 1)x + 2a + 6 = 0 ). The problem says it has one positive and one negative real root, and I need to find the range of the real number ( a ). Hmm, let me think about how to approach this.First, I remember that for a quadratic equation ( ax^2 + bx + c = 0 ), the roots can be positive, negative, or a mix. Since the problem mentions one positive and one negative root, I think this has to do with the product of the roots. I recall that the product of the roots of a quadratic equation is ( c/a ). In this case, the quadratic is ( x^2 + 2(a - 1)x + 2a + 6 ), so the product of the roots should be ( (2a + 6)/1 = 2a + 6 ).If one root is positive and the other is negative, their product must be negative because a positive times a negative is negative. So, the product of the roots ( 2a + 6 ) must be less than zero. That gives me the inequality ( 2a + 6 < 0 ). Solving for ( a ), I subtract 6 from both sides: ( 2a < -6 ), then divide by 2: ( a < -3 ). So, that suggests that ( a ) must be less than -3.But wait, I also remember that for a quadratic equation to have real roots, the discriminant must be non-negative. The discriminant ( D ) is ( b^2 - 4ac ). In this equation, ( a = 1 ), ( b = 2(a - 1) ), and ( c = 2a + 6 ). Let me compute the discriminant:( D = [2(a - 1)]^2 - 4 * 1 * (2a + 6) )First, square ( 2(a - 1) ): ( 4(a - 1)^2 )Then, expand ( (a - 1)^2 ): ( a^2 - 2a + 1 )So, ( 4(a^2 - 2a + 1) = 4a^2 - 8a + 4 )Now, subtract ( 4*(2a + 6) ): which is ( 8a + 24 )So, ( D = 4a^2 - 8a + 4 - 8a - 24 )Combine like terms: ( 4a^2 - 16a - 20 )For the equation to have real roots, the discriminant must be greater than or equal to zero:( 4a^2 - 16a - 20 geq 0 )Let me simplify this inequality. First, divide all terms by 4 to make it easier:( a^2 - 4a - 5 geq 0 )Now, factor the quadratic:Looking for two numbers that multiply to -5 and add to -4. Hmm, that would be -5 and +1.So, ( (a - 5)(a + 1) geq 0 )To solve this inequality, I can find the critical points where the expression equals zero: ( a = 5 ) and ( a = -1 ). These points divide the number line into three intervals:1. ( a < -1 )2. ( -1 < a < 5 )3. ( a > 5 )I need to test each interval to see where the product is non-negative.- For ( a < -1 ), let's pick ( a = -2 ): ( (-2 - 5)(-2 + 1) = (-7)(-1) = 7 ) which is positive. So, this interval satisfies the inequality.- For ( -1 < a < 5 ), let's pick ( a = 0 ): ( (0 - 5)(0 + 1) = (-5)(1) = -5 ) which is negative. So, this interval does not satisfy the inequality.- For ( a > 5 ), let's pick ( a = 6 ): ( (6 - 5)(6 + 1) = (1)(7) = 7 ) which is positive. So, this interval satisfies the inequality.Therefore, the discriminant is non-negative when ( a leq -1 ) or ( a geq 5 ). But earlier, I found that ( a < -3 ) for the product of the roots to be negative. So, I need to find the intersection of these two conditions.The discriminant condition gives ( a leq -1 ) or ( a geq 5 ), but the product condition requires ( a < -3 ). So, the overlap is ( a < -3 ) because ( a < -3 ) is already included in ( a leq -1 ). Therefore, the range of ( a ) is all real numbers less than -3.Wait, let me double-check. If ( a < -3 ), then it's definitely less than -1, so the discriminant is positive, and the product of the roots is negative, which means one positive and one negative root. That makes sense.Is there any other condition I might have missed? Maybe checking if the quadratic actually crosses the x-axis at two points, but since the discriminant is positive, it does have two real roots. And since the product is negative, they must be of opposite signs. So, I think that's all.So, putting it all together, the range of ( a ) is ( a < -3 ).**Final Answer**The range of the real number ( a ) is boxed{a < -3}.

💡Okay, so I have this problem about triangle ABC where angle BAC has a cosine of 4/5, and the triangle is isosceles with AB equal to AC. I need to find the cosine of angle BCA. Hmm, let me think about how to approach this.First, I know that in an isosceles triangle, two sides are equal, and the angles opposite those sides are also equal. Since AB equals AC, that means angles at B and C are equal. So, angle ABC equals angle BCA. Let me denote angle BAC as angle A, and angles ABC and BCA as angles B and C respectively. So, angle A is at vertex A, and angles B and C are at vertices B and C.Given that cos(angle A) is 4/5, I can find angle A using the arccos function. Let me calculate that:angle A = arccos(4/5)I can leave it like that for now, or maybe convert it to degrees if needed, but I think working in radians might be fine since I'm dealing with trigonometric functions.Since the triangle is isosceles with AB = AC, angles B and C are equal. Let me denote each of these angles as x. So, angle B = angle C = x.In any triangle, the sum of the angles is 180 degrees, or π radians. So, angle A + angle B + angle C = π.Substituting the known values:angle A + x + x = πangle A + 2x = πSo, 2x = π - angle Ax = (π - angle A)/2So, angle B and angle C are each equal to (π - angle A)/2.Now, I need to find cos(angle C), which is the same as cos(x). So, I need to find cos((π - angle A)/2).I remember that there's a trigonometric identity for cosine of a half-angle. The identity is:cos(θ/2) = ±√[(1 + cosθ)/2]But in this case, I have cos((π - angle A)/2). Let me see if I can relate this to the half-angle identity.Let me set θ = π - angle A. Then, cos((π - angle A)/2) = cos(θ/2) = ±√[(1 + cosθ)/2]But θ = π - angle A, so cosθ = cos(π - angle A). I also remember that cos(π - α) = -cosα. So, cosθ = -cos(angle A).Therefore, cos((π - angle A)/2) = ±√[(1 - cos(angle A))/2]Now, I need to determine the sign. Since angle A is between 0 and π (because it's an angle in a triangle), (π - angle A)/2 will be between 0 and π/2. So, the cosine of that angle will be positive. Therefore, I can drop the negative sign.So, cos(x) = √[(1 - cos(angle A))/2]Given that cos(angle A) = 4/5, let's plug that in:cos(x) = √[(1 - 4/5)/2] = √[(1/5)/2] = √(1/10) = √10 / 10Wait, but that's positive. However, in the initial problem, the triangle is isosceles with AB = AC, so angle BAC is at vertex A, and angles at B and C are equal. But in the initial problem statement, the user mentioned that the answer was negative. That seems conflicting.Wait, maybe I made a mistake in the angle identification. Let me double-check.If AB = AC, then sides AB and AC are equal, so the base is BC, and the two equal sides are AB and AC. Therefore, angles at B and C are equal. So, angle B and angle C are equal.But in the initial problem, the user's solution had a negative cosine. That suggests that angle C is obtuse, but in an isosceles triangle with two equal sides, the base angles are equal and acute if the vertex angle is acute.Wait, let's see: if cos(angle A) is 4/5, then angle A is acute because cosine is positive. So, angle A is less than 90 degrees. Therefore, angles B and C must each be (180 - angle A)/2, which would also be acute angles. Therefore, their cosines should be positive.But in the initial solution, the user got a negative cosine. That seems contradictory. Maybe I need to check the steps again.Wait, perhaps I confused the angles. Let me clarify:In triangle ABC, AB = AC, so it's an isosceles triangle with AB = AC. Therefore, the base is BC, and the two equal sides are AB and AC. Therefore, angles at B and C are equal.Given that, angle BAC is the vertex angle, which is given with cos(angle BAC) = 4/5. So, angle BAC is acute, as cosine is positive.Therefore, angles at B and C are equal and acute, so their cosines should be positive.But in the initial solution, the user had a negative cosine. That suggests that maybe the triangle was considered with AB = BC instead of AB = AC, which would change the configuration.Wait, let me check the problem statement again: "triangle ABC is an isosceles triangle with AB = AC." So, AB = AC, so the base is BC, and angles at B and C are equal.Therefore, angles B and C are equal and acute, so their cosines should be positive.But in the initial solution, the user had a negative cosine. That suggests that perhaps the user considered angle BAC as the base angle, which would be incorrect.Alternatively, maybe the user made a mistake in the angle identification.Wait, perhaps the user used a different approach, considering the triangle with sides AB = AC, but then using the law of cosines or something else, leading to a negative cosine.Wait, let me try using the law of cosines to find the sides and then find the cosine of angle C.Let me denote AB = AC = c, BC = a, and angle BAC = A.Given that cos(A) = 4/5, so sin(A) would be sqrt(1 - (16/25)) = 3/5.Using the law of cosines on angle A:a² = b² + c² - 2bc cos(A)But since AB = AC = c, and BC = a, so:a² = c² + c² - 2c²*(4/5) = 2c² - (8/5)c² = (10c² - 8c²)/5 = (2c²)/5So, a = c*sqrt(2/5)Now, to find angle C, which is equal to angle B.Using the law of cosines again for angle C:cos(C) = (a² + b² - c²)/(2ab)But in this case, sides opposite angle C is AB, which is c, and sides adjacent are AC = c and BC = a.Wait, no, in the law of cosines, for angle C, the sides are:cos(C) = (AC² + BC² - AB²)/(2*AC*BC)But AC = AB = c, and BC = a.So, cos(C) = (c² + a² - c²)/(2*c*a) = (a²)/(2*c*a) = a/(2c)From earlier, a = c*sqrt(2/5), so:cos(C) = (c*sqrt(2/5))/(2c) = sqrt(2/5)/2 = sqrt(2)/sqrt(5)/2 = sqrt(2)/(2*sqrt(5)) = sqrt(10)/10Which is positive, as expected.So, cos(angle C) = sqrt(10)/10, which is approximately 0.316.But in the initial solution, the user had a negative cosine, which contradicts this result.Wait, perhaps the user considered angle BAC as the base angle, which would be incorrect because in an isosceles triangle with AB = AC, angle BAC is the vertex angle, not the base angle.Alternatively, maybe the user used a different approach, such as considering the triangle with AB = BC instead of AB = AC, which would change the configuration.Wait, let me check the problem statement again: "triangle ABC is an isosceles triangle with AB = AC." So, AB = AC, so the base is BC, and angles at B and C are equal.Therefore, the correct approach is to find cos(angle C) as sqrt(10)/10, which is positive.But in the initial solution, the user had a negative cosine, which suggests a mistake in their approach.Alternatively, perhaps the user used the law of sines incorrectly, leading to a negative cosine.Wait, let me try using the law of sines.Given that AB = AC = c, angle BAC = A, angles at B and C = x.From the law of sines:a/sin(A) = c/sin(x)But a = BC, which we found earlier as c*sqrt(2/5)So,c*sqrt(2/5)/sin(A) = c/sin(x)Simplify:sqrt(2/5)/sin(A) = 1/sin(x)Therefore,sin(x) = sin(A)/sqrt(2/5)We know that sin(A) = 3/5, so:sin(x) = (3/5)/sqrt(2/5) = (3/5) * sqrt(5/2) = 3/(5) * sqrt(5)/sqrt(2) = 3*sqrt(5)/(5*sqrt(2)) = 3*sqrt(10)/10So, sin(x) = 3*sqrt(10)/10Now, to find cos(x), we can use the identity cos²(x) + sin²(x) = 1So,cos²(x) = 1 - sin²(x) = 1 - (9*10)/100 = 1 - 90/100 = 10/100 = 1/10Therefore, cos(x) = sqrt(1/10) = sqrt(10)/10, which is positive, as expected.So, cos(angle C) = sqrt(10)/10.Therefore, the initial solution's negative cosine was incorrect.Alternatively, perhaps the user considered angle BAC as the base angle, leading to a different configuration.Wait, if AB = AC, then angle BAC is the vertex angle, and angles at B and C are the base angles, which are equal.If instead, AB = BC, then angle BAC would be a base angle, and angles at B and C would be different.But in the problem statement, it's specified that AB = AC, so the base is BC, and angles at B and C are equal.Therefore, the correct value of cos(angle C) is sqrt(10)/10.Wait, but in the initial problem, the user's solution had a negative cosine, which suggests that perhaps they considered angle BAC as the base angle, leading to a different result.Alternatively, perhaps the user made a mistake in the angle identification or in the trigonometric identities.In any case, based on the correct approach, considering AB = AC, angle BAC is the vertex angle, and angles at B and C are equal and acute, so their cosines are positive.Therefore, the correct value of cos(angle BCA) is sqrt(10)/10.But wait, let me double-check the calculations.From the law of cosines:a² = 2c² - 2c²*(4/5) = 2c² - (8/5)c² = (10c² - 8c²)/5 = 2c²/5So, a = c*sqrt(2/5)Then, using the law of cosines for angle C:cos(C) = (c² + a² - c²)/(2*c*a) = a²/(2*c*a) = a/(2c) = (c*sqrt(2/5))/(2c) = sqrt(2/5)/2 = sqrt(10)/10Yes, that's correct.Alternatively, using the law of sines:sin(x) = 3*sqrt(10)/10Then, cos(x) = sqrt(1 - (9*10)/100) = sqrt(1 - 90/100) = sqrt(10)/10So, both methods confirm that cos(angle C) = sqrt(10)/10.Therefore, the initial solution's negative cosine was incorrect, likely due to a misapplication of the trigonometric identities or incorrect angle identification.So, the correct answer should be sqrt(10)/10.

💡Okay, so I have this problem where I need to find the range of values for ( b ) given that ( a ), ( b ), and ( c ) are real numbers, ( ac = b^2 ), and ( a + b + c = 3 ). The options are given as A: [0, 1], B: [-3, -1], C: [-1, 1], and D: [-3, 1]. Hmm, let me try to figure this out step by step.First, I know that ( ac = b^2 ). That seems like a quadratic relationship. Also, the sum ( a + b + c = 3 ). Maybe I can express ( a ) and ( c ) in terms of ( b ) somehow.Let me write down the equations:1. ( a + b + c = 3 )2. ( ac = b^2 )From the first equation, if I subtract ( b ) from both sides, I get ( a + c = 3 - b ). So, ( a + c = 3 - b ) and ( ac = b^2 ). Hmm, this looks like the sum and product of two numbers, which reminds me of quadratic equations. If I think of ( a ) and ( c ) as roots of a quadratic equation, then the quadratic would be ( x^2 - (a + c)x + ac = 0 ).Substituting the known values, that would be ( x^2 - (3 - b)x + b^2 = 0 ). So, the quadratic equation is ( x^2 - (3 - b)x + b^2 = 0 ).Now, since ( a ) and ( c ) are real numbers, the discriminant of this quadratic must be non-negative. The discriminant ( D ) is given by ( D = [-(3 - b)]^2 - 4 times 1 times b^2 ).Let me compute that:( D = (3 - b)^2 - 4b^2 )Expanding ( (3 - b)^2 ):( (3 - b)^2 = 9 - 6b + b^2 )So, substituting back into ( D ):( D = 9 - 6b + b^2 - 4b^2 )Combine like terms:( D = 9 - 6b + b^2 - 4b^2 = 9 - 6b - 3b^2 )So, ( D = -3b^2 - 6b + 9 ). For the quadratic to have real roots, the discriminant must be greater than or equal to zero:( -3b^2 - 6b + 9 geq 0 )Hmm, this is a quadratic inequality. Let me rearrange it to make it easier to solve. I can factor out a -3:( -3(b^2 + 2b - 3) geq 0 )Divide both sides by -3, remembering that dividing by a negative number reverses the inequality sign:( b^2 + 2b - 3 leq 0 )Now, let's solve the quadratic inequality ( b^2 + 2b - 3 leq 0 ). First, find the roots of the equation ( b^2 + 2b - 3 = 0 ).Using the quadratic formula:( b = frac{-2 pm sqrt{(2)^2 - 4 times 1 times (-3)}}{2 times 1} )Calculate the discriminant:( sqrt{4 + 12} = sqrt{16} = 4 )So, the roots are:( b = frac{-2 + 4}{2} = 1 ) and ( b = frac{-2 - 4}{2} = -3 )So, the quadratic ( b^2 + 2b - 3 ) factors as ( (b - 1)(b + 3) ). The inequality is ( (b - 1)(b + 3) leq 0 ).To solve this inequality, let's consider the intervals determined by the roots ( b = -3 ) and ( b = 1 ):1. When ( b < -3 ): Let's pick ( b = -4 ). Then, ( (-4 - 1)(-4 + 3) = (-5)(-1) = 5 ), which is positive. So, the inequality is not satisfied here.2. When ( -3 < b < 1 ): Let's pick ( b = 0 ). Then, ( (0 - 1)(0 + 3) = (-1)(3) = -3 ), which is negative. So, the inequality is satisfied here.3. When ( b > 1 ): Let's pick ( b = 2 ). Then, ( (2 - 1)(2 + 3) = (1)(5) = 5 ), which is positive. So, the inequality is not satisfied here.Additionally, at the critical points ( b = -3 ) and ( b = 1 ), the expression equals zero, which satisfies the inequality ( leq 0 ).Therefore, the solution to the inequality is ( -3 leq b leq 1 ).So, the range of values for ( b ) is from -3 to 1, which corresponds to option D: [-3, 1].Wait, just to make sure I didn't make any mistakes. Let me double-check my steps.1. I started with the given equations ( a + b + c = 3 ) and ( ac = b^2 ).2. Expressed ( a + c = 3 - b ) and recognized that ( a ) and ( c ) are roots of the quadratic ( x^2 - (3 - b)x + b^2 = 0 ).3. Calculated the discriminant ( D = (3 - b)^2 - 4b^2 = 9 - 6b + b^2 - 4b^2 = 9 - 6b - 3b^2 ).4. Set ( D geq 0 ) leading to ( -3b^2 - 6b + 9 geq 0 ), which simplifies to ( b^2 + 2b - 3 leq 0 ).5. Solved the quadratic inequality, found roots at ( b = 1 ) and ( b = -3 ), tested intervals, and concluded ( -3 leq b leq 1 ).Everything seems to check out. So, I think option D is correct.**Final Answer**boxed{D}