If the roots of the equation ax^2 + bx + c = | vedclass

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(A) Let the roots be $x_1 = \frac{\alpha}{\alpha - 1}$ and $x_2 = \frac{\alpha + 1}{\alpha}$.From the sum of roots: $x_1 + x_2 = \frac{\alpha}{\alpha

Vedclass · Accepted Answer

$b^2 - 4ac$

Vedclass · Answer

(A) Let the roots be $x_1 = \frac{\alpha}{\alpha - 1}$ and $x_2 = \frac{\alpha + 1}{\alpha}$.From the sum of roots: $x_1 + x_2 = \frac{\alpha}{\alpha - 1} + \frac{\alpha + 1}{\alpha} = \frac{\alpha^2 + \alpha^2 - 1}{\alpha(\alpha - 1)} = \frac{2\alpha^2 - 1}{\alpha^2 - \alpha} = -\frac{b}{a}$.From the product of roots: $x_1 x_2 = \frac{\alpha}{\alpha - 1} \cdot \frac{\alpha + 1}{\alpha} = \frac{\alpha + 1}{\alpha - 1} = \frac{c}{a}$.From $\frac{\alpha + 1}{\alpha - 1} = \frac{c}{a}$,we apply componendo and dividendo: $\frac{(\alpha + 1) + (\alpha - 1)}{(\alpha + 1) - (\alpha - 1)} = \frac{c + a}{c - a} \Rightarrow \frac{2\alpha}{2} = \frac{c + a}{c - a} \Rightarrow \alpha = \frac{c + a}{c - a}$.Also,$\alpha - 1 = \frac{c + a}{c - a} - 1 = \frac{2a}{c - a}$.Substituting these into the sum of roots equation: $\frac{2\alpha^2 - 1}{\alpha(\alpha - 1)} = -\frac{b}{a}$.After algebraic simplification,we find that $(a + b + c)^2 = b^2 - 4ac$.

If the roots of the equation $ax^2 + bx + c = 0$ are real and of the form $\frac{\alpha}{\alpha - 1}$ and $\frac{\alpha + 1}{\alpha}$,then the value of $(a + b + c)^2$ is

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