Let a, b in mathbb{R} be such that the equati | vedclass

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(B) For the equation $ax^{2}-2bx+15=0$,since it has a repeated root $\alpha$,the discriminant must be zero:$D = (-2b)^{2} - 4(a)(15) = 0 \implies 4b^{

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$58$

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(B) For the equation $ax^{2}-2bx+15=0$,since it has a repeated root $\alpha$,the discriminant must be zero:$D = (-2b)^{2} - 4(a)(15) = 0 \implies 4b^{2} = 60a \implies b^{2} = 15a$.Also,the root $\alpha = -\frac{-2b}{2a} = \frac{b}{a}$.Substituting $a = \frac{b^{2}}{15}$ into $\alpha = \frac{b}{a}$,we get $\alpha = \frac{b}{b^{2}/15} = \frac{15}{b}$.Since $\alpha$ is a root of $x^{2}-2bx+21=0$,we have:$(\frac{15}{b})^{2} - 2b(\frac{15}{b}) + 21 = 0$$\frac{225}{b^{2}} - 30 + 21 = 0 \implies \frac{225}{b^{2}} = 9 \implies b^{2} = 25$.For the equation $x^{2}-2bx+21=0$,the sum of roots $\alpha+\beta = 2b$ and the product $\alpha\beta = 21$.We need to find $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2} - 2\alpha\beta$.$\alpha^{2}+\beta^{2} = (2b)^{2} - 2(21) = 4b^{2} - 42$.Substituting $b^{2} = 25$:$\alpha^{2}+\beta^{2} = 4(25) - 42 = 100 - 42 = 58$.

Let $a, b \in \mathbb{R}$ be such that the equation $ax^{2}-2bx+15=0$ 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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