"Let a, b $\in$ R be such that the equation $a{x^2} - 2bx + 15 = 0$ has a repeated root $\alpha$. If $\alpha$ and $\beta$ are the roots of the equation ${x^2} - 2bx + 21 = 0$, then ${\alpha ^2} + {\beta ^2}$ is equal to :"

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Accepted Answer

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$a{x^2} - 2bx + 15 = 0$ has repeated root so ${b^2} = 15a$ and $\alpha = {{15} \over b}$

$\because$ $\alpha$ is a root of ${x^2} - 2bx + 21 = 0$

So ${{225} \over {{b^2}}} = 9 \Rightarrow {b^2} = 25$

Now $${\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta = 4{b^2} - 42 = 100 - 42 = 58$$

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Let a, b $\in$ R be such that the equation $a{x^2} - 2bx + 15 = 0$ has a repeated root $\alpha$. If $\alpha$ and $\beta$ are the roots of the equation ${x^2} - 2bx + 21 = 0$, then ${\alpha ^2} + {\beta ^2}$ is equal to :

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Let a, b $\in$ R be such that the equation $a{x^2} - 2bx + 15 = 0$ has a repeated root $\alpha$. If $\alpha$ and $\beta$ are the roots of the equation ${x^2} - 2bx + 21 = 0$, then ${\alpha ^2} + {\beta ^2}$ is equal to :

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