If (x_1, y_1) are the roots of x^2 + 8x - 20 | vedclass

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(A) For $x^2 + 8x - 20 = 0$,the roots are $x = 2, -10$. Thus,$(x_1, y_1) = (2, -10)$.For $4x^2 + 32x - 57 = 0$,the roots are $x = \frac{3}{2}, -\frac{

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are collinear

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(A) For $x^2 + 8x - 20 = 0$,the roots are $x = 2, -10$. Thus,$(x_1, y_1) = (2, -10)$.For $4x^2 + 32x - 57 = 0$,the roots are $x = \frac{3}{2}, -\frac{19}{2}$. Thus,$(x_2, y_2) = (\frac{3}{2}, -\frac{19}{2})$.For $9x^2 + 72x - 112 = 0$,the roots are $x = \frac{4}{3}, -\frac{28}{3}$. Thus,$(x_3, y_3) = (\frac{4}{3}, -\frac{28}{3})$.To check for collinearity,we find the equation of the line passing through $A(2, -10)$ and $B(\frac{3}{2}, -\frac{19}{2})$.Slope $m = \frac{-\frac{19}{2} - (-10)}{\frac{3}{2} - 2} = \frac{0.5}{-0.5} = -1$.Equation: $y - (-10) = -1(x - 2) \implies y + 10 = -x + 2 \implies x + y = -8$.Now,check if $C(\frac{4}{3}, -\frac{28}{3})$ satisfies $x + y = -8$:$\frac{4}{3} + (-\frac{28}{3}) = -\frac{24}{3} = -8$.Since the point $C$ satisfies the equation,the points are collinear.

If $(x_1, y_1)$ are the roots of $x^2 + 8x - 20 = 0$,$(x_2, y_2)$ are the roots of $4x^2 + 32x - 57 = 0$ and $(x_3, y_3)$ are the roots of $9x^2 + 72x - 112 = 0$,then the points $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$:

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