💡Alright, so I've got this problem here about quadratic equations with integer coefficients. It says that if such an equation has a rational root, then at least one of the coefficients ( a ), ( b ), or ( c ) must be even. Hmm, okay, let me try to wrap my head around this.First, I remember that for a quadratic equation ( ax^2 + bx + c = 0 ), the roots can be found using the quadratic formula: ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ). Since the equation has integer coefficients, the discriminant ( b^2 - 4ac ) must be a perfect square for the roots to be rational. That makes sense because the square root of a perfect square is an integer, which would make the entire expression rational.So, if the discriminant is a perfect square, let's say ( k^2 ), then we have ( b^2 - 4ac = k^2 ). Rearranging this, we get ( b^2 - k^2 = 4ac ). This looks like a difference of squares, which can be factored as ( (b - k)(b + k) = 4ac ).Now, since ( a ), ( b ), ( c ), and ( k ) are all integers, both ( b - k ) and ( b + k ) must be even or both must be odd. But wait, if ( b ) and ( k ) are both even or both odd, their sum and difference will be even. Let me check that:- If ( b ) is even and ( k ) is even, then ( b - k ) and ( b + k ) are both even.- If ( b ) is odd and ( k ) is odd, then ( b - k ) and ( b + k ) are both even.- If one is even and the other is odd, then ( b - k ) and ( b + k ) would both be odd.But in the case where one is even and the other is odd, their product ( (b - k)(b + k) ) would be odd times odd, which is odd. However, the right side of the equation is ( 4ac ), which is clearly even because it's a multiple of 4. So, if ( (b - k)(b + k) ) is odd, it can't equal ( 4ac ), which is even. Therefore, ( b ) and ( k ) must both be even or both be odd.But wait, if both ( b ) and ( k ) are even, then ( b ) is even. If both are odd, then ( b ) is odd. So, does this mean that ( b ) must be even? Not necessarily, because ( b ) could be odd if ( k ) is also odd. Hmm, so maybe I need to look deeper into the factors.Given that ( (b - k)(b + k) = 4ac ), and both ( b - k ) and ( b + k ) are even, let's denote ( b - k = 2m ) and ( b + k = 2n ), where ( m ) and ( n ) are integers. Then, substituting back, we have:( 2m times 2n = 4ac ) ( 4mn = 4ac ) Dividing both sides by 4: ( mn = ac )So, ( mn = ac ). Now, ( m ) and ( n ) are integers, and ( a ), ( c ) are integers as well. So, this tells us that ( a ) and ( c ) must multiply to give ( mn ). But how does this help us determine if at least one of ( a ), ( b ), or ( c ) is even?Well, let's think about the parity of these variables. If ( b ) is even, then we're done because ( b ) is even. If ( b ) is odd, then ( k ) must also be odd, as we established earlier. So, if ( b ) is odd, then ( k ) is odd, which means ( b - k ) and ( b + k ) are both even, as we saw. So, in this case, ( m ) and ( n ) are integers such that ( 2m ) and ( 2n ) multiply to ( 4ac ).But since ( mn = ac ), if both ( m ) and ( n ) are integers, then ( a ) and ( c ) must be such that their product is an integer. But we already know that ( a ) and ( c ) are integers, so that doesn't add new information.Wait, maybe I need to consider the factors of 4 in the equation ( (b - k)(b + k) = 4ac ). Since ( b - k ) and ( b + k ) are both even, their product is divisible by 4. So, ( 4ac ) is divisible by 4, which it obviously is. But does this tell us anything about the parity of ( a ), ( b ), or ( c )?If ( b ) is even, then we're done. If ( b ) is odd, then ( k ) is also odd, so ( b - k ) and ( b + k ) are both even, but not necessarily divisible by 4. Wait, no, because ( b ) and ( k ) are both odd, their difference and sum are even, but not necessarily multiples of 4. For example, if ( b = 3 ) and ( k = 1 ), then ( b - k = 2 ) and ( b + k = 4 ). So, in this case, one factor is 2 and the other is 4, which still gives a product divisible by 4.But in this example, ( a ) and ( c ) would have to multiply to ( mn = (2)(4)/4 = 2 ). So, ( a ) and ( c ) could be 1 and 2, for example. In this case, ( a = 1 ) is odd, ( c = 2 ) is even, so at least one of them is even.Wait, so if ( b ) is odd, then ( a ) or ( c ) must be even? Because in the example above, ( a ) was odd and ( c ) was even. Is that always the case?Let me test another example. Suppose ( b = 5 ) and ( k = 3 ). Then, ( b - k = 2 ) and ( b + k = 8 ). So, ( mn = (2)(8)/4 = 4 ). So, ( a ) and ( c ) could be 1 and 4, or 2 and 2, etc. If ( a = 1 ) and ( c = 4 ), then ( c ) is even. If ( a = 2 ) and ( c = 2 ), both are even. So, in either case, at least one of ( a ) or ( c ) is even.Another example: ( b = 7 ), ( k = 1 ). Then, ( b - k = 6 ), ( b + k = 8 ). So, ( mn = (6)(8)/4 = 12 ). So, ( a ) and ( c ) could be 3 and 4, or 2 and 6, etc. Again, at least one of them is even.Wait, is it possible for both ( a ) and ( c ) to be odd when ( b ) is odd? Let's see. Suppose ( a = 1 ) and ( c = 1 ). Then, ( ac = 1 ), so ( mn = 1 ). But ( mn = 1 ) implies that ( m = 1 ) and ( n = 1 ), so ( b - k = 2 ) and ( b + k = 2 ). This would mean ( b = 2 ) and ( k = 0 ), but ( k ) must be odd if ( b ) is odd, which contradicts ( k = 0 ). So, it's impossible for both ( a ) and ( c ) to be odd when ( b ) is odd.Therefore, if ( b ) is odd, at least one of ( a ) or ( c ) must be even. And if ( b ) is even, then we already have an even coefficient. So, in either case, at least one of ( a ), ( b ), or ( c ) must be even.Wait, but the problem statement says "at least one of ( a ), ( b ), or ( c ) is an even number." So, does this mean that if the equation has a rational root, then one of them must be even? Yes, based on the above reasoning.But let me try to formalize this a bit more. Suppose, for contradiction, that all of ( a ), ( b ), and ( c ) are odd. Then, let's see what happens.If ( a ), ( b ), and ( c ) are all odd, then the discriminant ( b^2 - 4ac ) would be ( text{odd}^2 - 4 times text{odd} times text{odd} ). Since ( text{odd}^2 ) is odd, and ( 4 times text{odd} times text{odd} ) is even (because 4 is even), so the discriminant would be ( text{odd} - text{even} = text{odd} ). But for the roots to be rational, the discriminant must be a perfect square, which is either even or odd. However, if the discriminant is odd, then ( k^2 ) is odd, so ( k ) must be odd.But earlier, we saw that if ( b ) is odd, then ( k ) must be odd, leading to ( b - k ) and ( b + k ) being even, which implies that ( a ) or ( c ) must be even. But in this case, we assumed all are odd, leading to a contradiction. Therefore, our assumption that all of ( a ), ( b ), and ( c ) are odd must be false. Hence, at least one of them must be even.So, to summarize:1. If the quadratic equation has a rational root, the discriminant must be a perfect square.2. This leads to the equation ( (b - k)(b + k) = 4ac ).3. If ( b ) is even, we're done.4. If ( b ) is odd, then ( k ) must also be odd, leading to ( a ) or ( c ) being even.5. Assuming all coefficients are odd leads to a contradiction, confirming that at least one must be even.Therefore, the statement is proven.

💡First, I need to determine how long it takes the first person to complete the work alone. Let's denote the number of days the first person takes to finish the work alone as ( x ).Prakash can complete the work in 30 days, so his work rate is ( frac{1}{30} ) of the work per day.When working together, they complete the work in 10 days, which means their combined work rate is ( frac{1}{10} ) of the work per day.The combined work rate is the sum of their individual work rates:[frac{1}{x} + frac{1}{30} = frac{1}{10}]To solve for ( x ), I'll find a common denominator for the fractions, which is ( 30x ):[frac{30}{30x} + frac{x}{30x} = frac{1}{10}]Multiplying every term by ( 30x ) to eliminate the denominators:[30 + x = 3x]Subtracting ( x ) from both sides to isolate ( x ):[30 = 2x]Finally, dividing both sides by 2 to solve for ( x ):[x = 15]Therefore, the first person can complete the work alone in 15 days.

💡Okay, so I have this trigonometry problem here: Given that sin(α + π/4) = 3/5, and α is between π/4 and 3π/4, I need to find cos α. Hmm, let me think about how to approach this.First, I remember that when dealing with angles added together inside a sine or cosine function, I can use the sine and cosine addition formulas. The problem gives me sin(α + π/4), so maybe I can expand that using the sine addition formula. The formula is sin(A + B) = sin A cos B + cos A sin B. So, applying that here:sin(α + π/4) = sin α cos(π/4) + cos α sin(π/4).I know that cos(π/4) and sin(π/4) are both √2/2. So substituting those in:sin(α + π/4) = sin α * (√2/2) + cos α * (√2/2).And this is equal to 3/5. So:(√2/2)(sin α + cos α) = 3/5.Hmm, okay, so if I let’s call sin α + cos α = x, then:(√2/2) x = 3/5.So, x = (3/5) * (2/√2) = (6)/(5√2). Maybe rationalizing the denominator:x = (6√2)/(5*2) = (3√2)/5.So, sin α + cos α = 3√2/5.Alright, that's one equation. But I need another equation to solve for cos α. I know that sin² α + cos² α = 1. So, maybe I can set up a system of equations.Let me denote sin α = s and cos α = c. Then:s + c = 3√2/5,ands² + c² = 1.I can solve this system for s and c. Let me square the first equation:(s + c)² = (3√2/5)²,which gives:s² + 2sc + c² = (9*2)/25 = 18/25.But I also know that s² + c² = 1, so substituting that in:1 + 2sc = 18/25.So, 2sc = 18/25 - 1 = 18/25 - 25/25 = -7/25.Therefore, sc = -7/50.So, now I have:s + c = 3√2/5,andsc = -7/50.This is a system of equations that can be solved by recognizing that s and c are roots of the quadratic equation:x² - (s + c)x + sc = 0,which becomes:x² - (3√2/5)x - 7/50 = 0.Let me write that down:x² - (3√2/5)x - 7/50 = 0.To solve this quadratic, I can use the quadratic formula:x = [ (3√2/5) ± sqrt( (3√2/5)^2 + 4*(7/50) ) ] / 2.First, compute the discriminant:D = (3√2/5)^2 + 4*(7/50).Calculating each term:(3√2/5)^2 = 9*2 / 25 = 18/25,and 4*(7/50) = 28/50 = 14/25.So, D = 18/25 + 14/25 = 32/25.Therefore, sqrt(D) = sqrt(32/25) = (4√2)/5.So, plugging back into the quadratic formula:x = [ (3√2/5) ± (4√2)/5 ] / 2.Let me compute both possibilities:First, with the plus sign:x = [ (3√2 + 4√2)/5 ] / 2 = [7√2/5] / 2 = 7√2/10.Second, with the minus sign:x = [ (3√2 - 4√2)/5 ] / 2 = [ (-√2)/5 ] / 2 = -√2/10.So, the solutions are x = 7√2/10 and x = -√2/10.Therefore, sin α and cos α are 7√2/10 and -√2/10 in some order.But which one is which? I need to figure out whether sin α is 7√2/10 and cos α is -√2/10, or vice versa.Given that α is between π/4 and 3π/4, which is the second quadrant. In the second quadrant, sine is positive and cosine is negative. So, sin α is positive, and cos α is negative.Therefore, sin α must be 7√2/10, and cos α must be -√2/10.Wait, but hold on. Let me check if these values make sense.Compute sin α + cos α: 7√2/10 + (-√2/10) = 6√2/10 = 3√2/5, which matches our earlier result.Also, sin² α + cos² α: (7√2/10)^2 + (-√2/10)^2 = (49*2)/100 + (2)/100 = 98/100 + 2/100 = 100/100 = 1. That checks out.So, yes, cos α is -√2/10.But wait, let me think again. The angle α is between π/4 and 3π/4, so it's in the second quadrant. So, cos α is negative, which matches our result.But let me double-check if I didn't make a mistake in the quadratic solution.We had:x² - (3√2/5)x - 7/50 = 0.Using quadratic formula:x = [ (3√2/5) ± sqrt( (3√2/5)^2 + 4*(7/50) ) ] / 2.Computed discriminant correctly as 32/25, sqrt(32/25) = 4√2/5.So, x = [3√2/5 ± 4√2/5]/2.Which gives:(3√2 + 4√2)/10 = 7√2/10,and(3√2 - 4√2)/10 = (-√2)/10.Yes, that seems correct.So, since sin α is positive and cos α is negative, cos α is -√2/10.Looking back at the options:A: 7√2/10B: -7√2/10C: √2/10D: -√2/10So, the answer should be D: -√2/10.Wait, but hold on a second. Let me think if there's another approach to verify this.Alternatively, since we know sin(α + π/4) = 3/5, and α is in (π/4, 3π/4), so α + π/4 is in (π/2, π). So, it's in the second quadrant where sine is positive and cosine is negative.So, maybe we can find cos(α + π/4) first.Using the identity sin²x + cos²x = 1,cos²(α + π/4) = 1 - sin²(α + π/4) = 1 - (9/25) = 16/25,so cos(α + π/4) = -4/5, since it's in the second quadrant.Then, using the cosine addition formula:cos α = cos[(α + π/4) - π/4] = cos(α + π/4)cos(π/4) + sin(α + π/4)sin(π/4).We know cos(α + π/4) = -4/5, sin(α + π/4) = 3/5, and cos(π/4) = sin(π/4) = √2/2.So,cos α = (-4/5)(√2/2) + (3/5)(√2/2) = [(-4 + 3)/5]*(√2/2) = (-1/5)(√2/2) = -√2/10.Yes, same result. So, that confirms it.Therefore, the answer is D: -√2/10.**Final Answer**boxed{D}

💡Okay, so I have this problem here: A line passes through the point (2,1) and intersects the coordinate axes at points A and B. The area of triangle BO (where O is the origin) is exactly 4. I need to find the number of lines l that satisfy these conditions.Hmm, let me try to visualize this. We have a coordinate plane with the origin O. The line passes through (2,1) and intersects the x-axis at point A and the y-axis at point B. The triangle formed by points B, O, and the origin is supposed to have an area of 4. So, triangle BO is actually triangle BAO? Wait, no, triangle BO would just be a triangle with vertices at B, O, and another point? Wait, maybe it's triangle AOB? Because A and B are on the axes, so triangle AOB would make sense with vertices at A, O, and B.But the problem says "the area of triangle BO," which is a bit confusing because BO is just a line segment. Maybe it's a typo or maybe I'm misinterpreting. Wait, let me check the original problem again. It says, "the area of triangle BO (where O is the origin) is exactly 4." Hmm, so maybe it's triangle BAO? Or perhaps it's triangle AOB? Maybe the problem meant triangle AOB. Because triangle BO wouldn't make sense as a triangle unless it's referring to triangle BAO or AOB.Wait, maybe it's triangle AOB. So, the area of triangle AOB is 4. That makes more sense because A and B are on the axes, so triangle AOB would be a right triangle with legs OA and OB. So, the area would be (1/2)*|OA|*|OB| = 4.Alright, so I think that's the correct interpretation. So, the area of triangle AOB is 4. Therefore, (1/2)*|OA|*|OB| = 4. So, |OA|*|OB| = 8.Now, the line passes through (2,1). Let me write the equation of the line in intercept form. The intercept form of a line is x/a + y/b = 1, where a is the x-intercept and b is the y-intercept. So, points A and B would be (a,0) and (0,b) respectively.Since the line passes through (2,1), substituting into the equation gives 2/a + 1/b = 1.Also, from the area condition, we have (1/2)*|a|*|b| = 4, so |a|*|b| = 8.So, we have two equations:1. 2/a + 1/b = 12. |a|*|b| = 8But since a and b are intercepts, they can be positive or negative, but the area is positive, so |a|*|b| = 8.But let's consider a and b as positive first, and then we can think about negative intercepts.So, if a and b are positive, then the line is in the first quadrant, passing through (2,1). If a is negative, the x-intercept is on the negative x-axis, and similarly, if b is negative, the y-intercept is on the negative y-axis.But let's first assume a and b are positive. So, we can drop the absolute value signs.So, we have:1. 2/a + 1/b = 12. a*b = 8From equation 2, we can express b = 8/a.Substitute into equation 1:2/a + 1/(8/a) = 1Simplify 1/(8/a) = a/8.So, equation becomes:2/a + a/8 = 1Multiply both sides by 8a to eliminate denominators:16 + a^2 = 8aBring all terms to one side:a^2 - 8a + 16 = 0This is a quadratic equation in a. Let's solve it:a = [8 ± sqrt(64 - 64)] / 2 = [8 ± 0]/2 = 4So, a = 4 is a repeated root. Therefore, b = 8/a = 8/4 = 2.So, in the case where a and b are positive, we have a unique solution: a=4, b=2.Therefore, the line is x/4 + y/2 = 1, which simplifies to x + 2y = 4.But wait, does this line pass through (2,1)? Let's check: 2 + 2*1 = 4, which is correct. So, yes, it does pass through (2,1).So, that's one line.But the problem says "the number of lines l that satisfy the given conditions." So, maybe there are more lines if we consider negative intercepts.So, let's consider cases where a or b is negative.Case 1: a positive, b negative.So, a > 0, b < 0.Then, equation 1: 2/a + 1/b = 1Equation 2: |a|*|b| = 8 => a*(-b) = 8 => -a*b = 8 => a*b = -8So, from equation 2: a*b = -8From equation 1: 2/a + 1/b = 1Express b from equation 2: b = -8/aSubstitute into equation 1:2/a + 1/(-8/a) = 1Simplify 1/(-8/a) = -a/8So, equation becomes:2/a - a/8 = 1Multiply both sides by 8a:16 - a^2 = 8aBring all terms to one side:-a^2 -8a +16 = 0Multiply both sides by -1:a^2 +8a -16 = 0Solve for a:a = [-8 ± sqrt(64 + 64)] / 2 = [-8 ± sqrt(128)] / 2 = [-8 ± 8*sqrt(2)] / 2 = -4 ± 4*sqrt(2)So, a = -4 + 4*sqrt(2) or a = -4 -4*sqrt(2)But since a is positive in this case, let's check:-4 + 4*sqrt(2) is approximately -4 + 5.656 = 1.656, which is positive.-4 -4*sqrt(2) is approximately -4 -5.656 = -9.656, which is negative, so we discard this.So, a = -4 + 4*sqrt(2) ≈1.656Then, b = -8/a ≈ -8 /1.656 ≈ -4.83So, this is a valid solution with a positive a and negative b.Therefore, another line exists in this case.Case 2: a negative, b positive.So, a < 0, b > 0.Equation 1: 2/a + 1/b = 1Equation 2: |a|*|b| = 8 => (-a)*b =8 => -a*b =8 => a*b = -8From equation 2: a*b = -8 => b = -8/aSubstitute into equation 1:2/a + 1/(-8/a) =1Simplify 1/(-8/a) = -a/8So, equation becomes:2/a - a/8 =1Multiply both sides by 8a:16 - a^2 =8aBring all terms to one side:-a^2 -8a +16 =0Multiply by -1:a^2 +8a -16=0Same as previous case.Solutions: a = [-8 ± sqrt(64 +64)] /2 = [-8 ± 8*sqrt(2)] /2 = -4 ±4*sqrt(2)Since a is negative in this case, let's check:-4 +4*sqrt(2) ≈-4 +5.656≈1.656, which is positive, so we discard.-4 -4*sqrt(2)≈-4 -5.656≈-9.656, which is negative, so we take this.Thus, a = -4 -4*sqrt(2)Then, b = -8/a = -8/(-4 -4*sqrt(2)) = 8/(4 +4*sqrt(2)) = 2/(1 +sqrt(2))Multiply numerator and denominator by (sqrt(2)-1):2*(sqrt(2)-1)/[(1 +sqrt(2))(sqrt(2)-1)] = 2*(sqrt(2)-1)/(2 -1) = 2*(sqrt(2)-1)So, b = 2*(sqrt(2)-1)≈2*(1.414 -1)=2*(0.414)=0.828So, this is a valid solution with a negative a and positive b.Therefore, another line exists in this case.Case 3: a negative, b negative.So, a <0, b<0.Equation 1: 2/a +1/b =1Equation 2: |a|*|b|=8 => (-a)*(-b)=8 => a*b=8So, from equation 2: a*b=8From equation 1: 2/a +1/b =1Express b from equation 2: b=8/aSubstitute into equation 1:2/a +1/(8/a)=1 => 2/a +a/8=1Multiply both sides by 8a:16 +a^2=8aBring all terms to one side:a^2 -8a +16=0Which is the same quadratic as before, leading to a=4, but a is negative in this case, so a=4 is positive, so no solution here.Therefore, no solution in this case.So, in total, we have three lines:1. The line with positive a and positive b: x/4 + y/2 =12. The line with positive a and negative b: x/( -4 +4*sqrt(2)) + y/( -8/( -4 +4*sqrt(2)) )=1, which simplifies to x/( -4 +4*sqrt(2)) + y/( 2*(sqrt(2)-1) )=13. The line with negative a and positive b: x/( -4 -4*sqrt(2)) + y/( 2*(sqrt(2)-1) )=1Wait, but let me check if these are distinct lines.Wait, in case 1, a=4, b=2.In case 2, a=-4 +4*sqrt(2), b= -8/a= -8/(-4 +4*sqrt(2))= 8/(4 -4*sqrt(2))= 2/(1 -sqrt(2))= multiply numerator and denominator by (1 +sqrt(2)): 2*(1 +sqrt(2))/(1 -2)= -2*(1 +sqrt(2))= -2 -2*sqrt(2). Wait, that contradicts my earlier calculation. Wait, no, let me recast.Wait, in case 2, a=-4 +4*sqrt(2), which is approximately 1.656, positive, so b= -8/a≈-4.83, which is negative.But in case 3, a is negative, so a=-4 -4*sqrt(2), which is approximately -9.656, and b=2*(sqrt(2)-1)≈0.828, positive.So, these are three distinct lines.But wait, in case 2 and case 3, are these lines different?Yes, because in case 2, a is positive, b is negative, and in case 3, a is negative, b is positive. So, these are two different lines.Therefore, in total, we have three lines: one with both intercepts positive, one with a positive and b negative, and one with a negative and b positive.Wait, but in case 3, when a is negative and b is positive, we have a unique solution, so that's another line.So, in total, three lines.But wait, let me check if the line in case 2 and case 3 are distinct.Yes, because in case 2, the x-intercept is positive, y-intercept is negative, and in case 3, x-intercept is negative, y-intercept is positive. So, these are two different lines.Therefore, in total, three lines satisfy the given conditions.Wait, but let me think again. When I considered case 2, a was positive, b negative, leading to one line. Case 3, a negative, b positive, leading to another line. And case 1, both positive, leading to the third line.So, total three lines.But wait, in case 2 and case 3, when I solved, I got two different lines, each with one positive and one negative intercept. So, that's two lines. Plus the one with both positive intercepts, total three.Therefore, the number of lines is three.But let me think again. Is there a possibility of more lines? For example, if the line is vertical or horizontal, but since it passes through (2,1), a vertical line would be x=2, which would intersect y-axis at (2,0), but then the area would be zero, which doesn't satisfy the area condition. Similarly, a horizontal line y=1 would intersect x-axis at (0,1), but again, the area would be zero. So, these are not valid.Therefore, only three lines satisfy the conditions.Wait, but in the initial solution, I considered the area as (1/2)*|OA|*|OB|=4, which led to |a|*|b|=8. But in the problem statement, it says "the area of triangle BO is exactly 4." Wait, triangle BO? That would be a triangle with vertices at B, O, and... Wait, triangle BO is just two points, so that doesn't make sense. Maybe it's a typo, and it should be triangle AOB.Alternatively, maybe it's triangle BAO, but that's the same as triangle AOB.Wait, perhaps the problem meant triangle ABO, which is the same as triangle AOB.So, I think my initial interpretation is correct, that the area is 4, leading to |a|*|b|=8.Therefore, with that, I think the number of lines is three.But wait, let me check my calculations again.In case 1: a=4, b=2, line x/4 + y/2=1, passes through (2,1): 2/4 +1/2=0.5 +0.5=1, correct.In case 2: a=-4 +4*sqrt(2), b= -8/a= -8/(-4 +4*sqrt(2))= 8/(4 -4*sqrt(2))= 2/(1 -sqrt(2))= multiply numerator and denominator by (1 +sqrt(2)): 2*(1 +sqrt(2))/(1 -2)= -2*(1 +sqrt(2))= -2 -2*sqrt(2). So, b= -2 -2*sqrt(2). So, the line is x/a + y/b=1, with a= -4 +4*sqrt(2), b= -2 -2*sqrt(2). Let's check if it passes through (2,1):2/a +1/b= 2/(-4 +4*sqrt(2)) +1/(-2 -2*sqrt(2)).Let me compute 2/(-4 +4*sqrt(2))= 2/(4(sqrt(2)-1))= (1)/(2(sqrt(2)-1))= multiply numerator and denominator by (sqrt(2)+1): (sqrt(2)+1)/(2*(2 -1))= (sqrt(2)+1)/2.Similarly, 1/(-2 -2*sqrt(2))= -1/(2 +2*sqrt(2))= -1/(2(1 +sqrt(2)))= multiply numerator and denominator by (sqrt(2)-1): - (sqrt(2)-1)/(2*(2 -1))= - (sqrt(2)-1)/2.So, total: (sqrt(2)+1)/2 - (sqrt(2)-1)/2= [sqrt(2)+1 -sqrt(2)+1]/2= 2/2=1. So, yes, it passes through (2,1).Similarly, in case 3: a= -4 -4*sqrt(2), b=2*(sqrt(2)-1). Let's check if the line passes through (2,1):2/a +1/b= 2/(-4 -4*sqrt(2)) +1/(2*(sqrt(2)-1)).Compute 2/(-4 -4*sqrt(2))= -2/(4 +4*sqrt(2))= -1/(2 +2*sqrt(2))= multiply numerator and denominator by (sqrt(2)-1): - (sqrt(2)-1)/(2*(2 -1))= - (sqrt(2)-1)/2.Compute 1/(2*(sqrt(2)-1))= multiply numerator and denominator by (sqrt(2)+1): (sqrt(2)+1)/(2*(2 -1))= (sqrt(2)+1)/2.So, total: - (sqrt(2)-1)/2 + (sqrt(2)+1)/2= [ -sqrt(2)+1 +sqrt(2)+1 ]/2= 2/2=1. So, yes, it passes through (2,1).Therefore, all three lines satisfy the conditions.Hence, the number of lines is three.

💡Okay, so I have this problem here where a function ( f ) is defined by ( fleft(frac{x}{3}right) = x^2 + x + 1 ). I need to find the sum of all values of ( z ) for which ( f(3z) = 4 ). The answer choices are given, so I can check my work against them once I figure it out.First, I need to understand what the function ( f ) does. It's defined in terms of ( frac{x}{3} ), which means that if I input ( frac{x}{3} ) into ( f ), I get ( x^2 + x + 1 ). So, to find ( f(3z) ), I need to figure out what ( x ) should be so that ( frac{x}{3} = 3z ). Let me write that down:If ( fleft(frac{x}{3}right) = x^2 + x + 1 ), then to find ( f(3z) ), I set ( frac{x}{3} = 3z ). Solving for ( x ), I multiply both sides by 3:( x = 9z ).So, substituting ( x = 9z ) into the function, we get:( f(3z) = (9z)^2 + 9z + 1 ).Let me compute that:( (9z)^2 = 81z^2 ),so ( f(3z) = 81z^2 + 9z + 1 ).Now, the problem states that ( f(3z) = 4 ). So, I set up the equation:( 81z^2 + 9z + 1 = 4 ).To solve for ( z ), I subtract 4 from both sides:( 81z^2 + 9z + 1 - 4 = 0 ),which simplifies to:( 81z^2 + 9z - 3 = 0 ).Now, I have a quadratic equation in terms of ( z ). The standard form of a quadratic equation is ( ax^2 + bx + c = 0 ), so here, ( a = 81 ), ( b = 9 ), and ( c = -3 ).I need to find the sum of all values of ( z ) that satisfy this equation. I remember that for a quadratic equation ( ax^2 + bx + c = 0 ), the sum of the roots is given by ( -frac{b}{a} ). So, applying that here:Sum of roots ( = -frac{9}{81} ).Simplifying that, ( -frac{9}{81} = -frac{1}{9} ).Wait, that seems straightforward. But let me double-check to make sure I didn't make a mistake. Maybe I should solve the quadratic equation explicitly to find the roots and then add them up.Using the quadratic formula, ( z = frac{-b pm sqrt{b^2 - 4ac}}{2a} ).Plugging in the values:( z = frac{-9 pm sqrt{9^2 - 4 cdot 81 cdot (-3)}}{2 cdot 81} ).Calculating the discriminant:( b^2 - 4ac = 81 - 4 cdot 81 cdot (-3) ).Wait, hold on. ( b^2 ) is ( 9^2 = 81 ). Then, ( 4ac = 4 cdot 81 cdot (-3) = 4 cdot (-243) = -972 ). So, the discriminant is ( 81 - (-972) = 81 + 972 = 1053 ).So, the square root of 1053 is... Hmm, let me see. 1053 divided by 9 is 117, which is 9 times 13. So, ( sqrt{1053} = sqrt{9 times 117} = 3sqrt{117} ). But ( 117 = 9 times 13 ), so ( sqrt{117} = 3sqrt{13} ). Therefore, ( sqrt{1053} = 3 times 3sqrt{13} = 9sqrt{13} ).So, plugging back into the quadratic formula:( z = frac{-9 pm 9sqrt{13}}{162} ).Simplifying numerator and denominator by dividing numerator and denominator by 9:( z = frac{-1 pm sqrt{13}}{18} ).So, the two roots are ( z = frac{-1 + sqrt{13}}{18} ) and ( z = frac{-1 - sqrt{13}}{18} ).Adding these two roots together:( frac{-1 + sqrt{13}}{18} + frac{-1 - sqrt{13}}{18} = frac{-1 + sqrt{13} -1 - sqrt{13}}{18} = frac{-2}{18} = -frac{1}{9} ).Okay, so that confirms it. The sum of the roots is indeed ( -frac{1}{9} ).Wait, but let me think again. The function ( f ) was defined as ( fleft(frac{x}{3}right) = x^2 + x + 1 ). So, when I set ( frac{x}{3} = 3z ), I got ( x = 9z ). Then, substituting back, I got ( f(3z) = 81z^2 + 9z + 1 ). That seems correct.Setting that equal to 4, I subtracted 4 and got the quadratic equation ( 81z^2 + 9z - 3 = 0 ). Then, using the sum of roots formula, I found ( -frac{1}{9} ). Solving the quadratic explicitly also gave me the same result. So, I think that's solid.But just to be absolutely thorough, let me check my substitution again. If ( fleft(frac{x}{3}right) = x^2 + x + 1 ), then to find ( f(y) ), where ( y = frac{x}{3} ), we can express ( x = 3y ). Therefore, ( f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y + 1 ).Ah, wait a second! Did I make a mistake earlier? Because if ( f(y) = 9y^2 + 3y + 1 ), then ( f(3z) = 9(3z)^2 + 3(3z) + 1 = 9 times 9z^2 + 9z + 1 = 81z^2 + 9z + 1 ). So, that's consistent with what I had before. So, no mistake there.Therefore, setting ( f(3z) = 4 ) gives ( 81z^2 + 9z + 1 = 4 ), leading to ( 81z^2 + 9z - 3 = 0 ). The sum of the roots is ( -frac{b}{a} = -frac{9}{81} = -frac{1}{9} ).So, the answer should be ( -frac{1}{9} ), which is option A.But just to make sure, let me consider if there's another way to approach this problem. Maybe by directly expressing ( f ) in terms of its argument.Given ( fleft(frac{x}{3}right) = x^2 + x + 1 ), let me let ( t = frac{x}{3} ). Then, ( x = 3t ), and substituting back, ( f(t) = (3t)^2 + 3t + 1 = 9t^2 + 3t + 1 ). So, ( f(t) = 9t^2 + 3t + 1 ).Therefore, ( f(3z) = 9(3z)^2 + 3(3z) + 1 = 9 times 9z^2 + 9z + 1 = 81z^2 + 9z + 1 ). Yep, same as before.Setting that equal to 4: ( 81z^2 + 9z + 1 = 4 ), so ( 81z^2 + 9z - 3 = 0 ). Sum of roots is ( -frac{1}{9} ).I think I've verified it multiple ways now, so I'm confident that the sum is ( -frac{1}{9} ).**Final Answer**boxed{A}

💡First, I need to determine the total number of students who received the same grade on both tests. I'll add up the number of students for each grade:- **A**: 3 students- **B**: 5 students- **C**: 6 students- **D**: 2 students- **E**: 1 studentAdding these together: 3 + 5 + 6 + 2 + 1 = 17 students.Next, I'll calculate the percentage of these 17 students out of the total 40 students in the class. To do this, I'll divide the number of students with the same grade by the total number of students and then multiply by 100 to get the percentage:[text{Percentage} = left( frac{17}{40} right) times 100% = 42.5%]Therefore, the percentage of students who received the same grades on both tests is 42.5%.