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If the sum of the rootsof the equation $a x^{2}+b x+c=0$ is equal to the sum of their squares, then
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The correct answer is:
$a b+b^{2}=2 a c$
$a \mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}=0$
Let the root be $\alpha$ and $\beta$.
$\alpha+\beta=-\frac{\mathrm{b}}{\mathrm{a}}, \alpha, \beta=\frac{\mathrm{c}}{\mathrm{a}}$
$\Rightarrow \alpha+\beta=\alpha^{2}+\beta^{2} \quad \ldots$ (given)
$\Rightarrow \alpha+\beta=(\alpha+\beta)^{2}-2 \alpha \beta$
$-\frac{\mathrm{b}}{\mathrm{a}}=\left(-\frac{\mathrm{b}}{\mathrm{a}}\right)^{2}-\frac{2 \mathrm{c}}{\mathrm{a}}$
$\Rightarrow \mathrm{b}^{2}+\mathrm{ab}=2 \mathrm{ac}$
Let the root be $\alpha$ and $\beta$.
$\alpha+\beta=-\frac{\mathrm{b}}{\mathrm{a}}, \alpha, \beta=\frac{\mathrm{c}}{\mathrm{a}}$
$\Rightarrow \alpha+\beta=\alpha^{2}+\beta^{2} \quad \ldots$ (given)
$\Rightarrow \alpha+\beta=(\alpha+\beta)^{2}-2 \alpha \beta$
$-\frac{\mathrm{b}}{\mathrm{a}}=\left(-\frac{\mathrm{b}}{\mathrm{a}}\right)^{2}-\frac{2 \mathrm{c}}{\mathrm{a}}$
$\Rightarrow \mathrm{b}^{2}+\mathrm{ab}=2 \mathrm{ac}$
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