Let a, b in R, a neq 0 be such that the equat | vedclass

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(B) For the equation $a x^{2}-2 b x+5=0$,since it has a repeated root $\alpha$,the discriminant $D = 0$.$D = (-2b)^{2} - 4(a)(5) = 4b^{2} - 20a = 0 \R

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$25$

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(B) For the equation $a x^{2}-2 b x+5=0$,since it has a repeated root $\alpha$,the discriminant $D = 0$.$D = (-2b)^{2} - 4(a)(5) = 4b^{2} - 20a = 0 \Rightarrow b^{2} = 5a$.The root $\alpha$ is given by $\alpha = -(-2b) / (2a) = b/a$.Since $\alpha$ is a root of $a x^{2}-2 b x+5=0$,we have $a(b/a)^{2} - 2b(b/a) + 5 = 0 \Rightarrow b^{2}/a - 2b^{2}/a + 5 = 0 \Rightarrow b^{2}/a = 5$,which is consistent with $b^{2} = 5a$.Given $\alpha$ is also a root of $x^{2}-2 b x-10=0$,we have $\alpha^{2} - 2b\alpha - 10 = 0$.Substituting $\alpha = b/a$,we get $(b/a)^{2} - 2b(b/a) - 10 = 0 \Rightarrow b^{2}/a^{2} - 2b^{2}/a - 10 = 0$.Since $b^{2} = 5a$,we substitute $b^{2}/a = 5$ and $b^{2} = 5a$ into the equation: $5/a - 2(5) - 10 = 0 \Rightarrow 5/a = 20 \Rightarrow a = 1/4$.Then $b^{2} = 5(1/4) = 5/4 \Rightarrow b = \pm \sqrt{5}/2$.Now,$\alpha = b/a = (\pm \sqrt{5}/2) / (1/4) = \pm 2\sqrt{5}$. Thus $\alpha^{2} = (\pm 2\sqrt{5})^{2} = 20$.For the equation $x^{2}-2bx-10=0$,the product of roots $\alpha\beta = -10$.So,$\beta = -10/\alpha = -10/(\pm 2\sqrt{5}) = \mp \sqrt{5}$.Thus $\beta^{2} = (\mp \sqrt{5})^{2} = 5$.Therefore,$\alpha^{2} + \beta^{2} = 20 + 5 = 25$.

Let $a, b \in R, a \neq 0$ 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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