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(B) For the equation $a x^{2}-2 b x+5=0$,the roots are $\alpha, \alpha$. Thus,the sum of roots $2\alpha = \frac{2b}{a} \Rightarrow \alpha = \frac{b}{a

Vedclass · Accepted Answer

$25$

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(B) For the equation $a x^{2}-2 b x+5=0$,the roots are $\alpha, \alpha$. Thus,the sum of roots $2\alpha = \frac{2b}{a} \Rightarrow \alpha = \frac{b}{a}$ and the product of roots $\alpha^{2} = \frac{5}{a}$. From these,$b = a\alpha$ and $a = \frac{5}{\alpha^{2}}$. Substituting $a$,we get $b = \frac{5}{\alpha}$. Since $\alpha$ is also a root of $x^{2}-2 b x-10=0$,we have $\alpha^{2}-2 b \alpha-10=0$. Substituting $b = \frac{5}{\alpha}$ into this equation: $\alpha^{2}-2(\frac{5}{\alpha})\alpha-10=0$ $\Rightarrow \alpha^{2}-10-10=0$ $\Rightarrow \alpha^{2}=20$. Now,for the equation $x^{2}-2 b x-10=0$,the product of roots $\alpha \beta = -10$. Since $\alpha^{2} = 20$,we have $\alpha = \pm \sqrt{20}$. Then $\beta = \frac{-10}{\alpha}$. Thus,$\beta^{2} = \frac{100}{\alpha^{2}} = \frac{100}{20} = 5$. Therefore,$\alpha^{2}+\beta^{2} = 20 + 5 = 25$.

Let $a, b \in \mathbb{R}, a \neq 0$ be such that the equation $a x^{2}-2 b x+5=0$ has a repeated root $\alpha,$ which is also a root of the equation $x^{2}-2 b x-10=0$. If $\beta$ is the other root of this equation,then $\alpha^{2}+\beta^{2}$ is equal to:

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