Let a, b $\in$ R, a $\ne$ 0 be such that the equation, ax² – 2bx + 5 = 0 has a repeated root $\alpha$, which is also a root of the equation, x² – 2bx – 10 = 0. If $\beta$ is the other root of this equation, then $\alpha$² + $\beta$² is equal to :

Roots of equation ax² – 2bx + 5 = 0 are $\alpha$, $\alpha$. <br><br>$\therefore$ $\alpha$ + $\alpha$ = ${{2b} \over a}$ <br><br>$\Rightarrow$ 2$\alpha$ = ${{2b} \over a}$ <br><br>$\Rightarrow$ $\alpha$ = ${{b} \over a}$ ....(1) <br><br>and $\alpha$² = ${5 \over a}$ ....(2) <br><br>From (1) and (2), we get <br><br>${{{b^2}} \over {{a^2}}} = {5 \over a}$ <br><br>$\Rightarrow$ b² = 5a <br><br>$\alpha$, $\beta$ are the roots of equation x² – 2bx – 10 = 0 <br><br>$\therefore$ $\alpha$ + $\beta$ = 2b <br><br>and $\alpha$$\beta$ = -10 <br><br>As $\alpha$ is a root of the equation x² – 2bx – 10 = 0. <br><br>$\therefore$ $\alpha$² - 2b$\alpha$ - 10 = 0 <br><br>$\Rightarrow$ ${{{b^2}} \over {{a^2}}} - {{2{b^2}} \over a} - 10 = 0$ <br><br>$\Rightarrow$ ${{5a} \over {{a^2}}} - {{10a} \over a} - 10 = 0$ <br><br>$\Rightarrow$ ${5 \over a} - 10 - 10 = 0$ <br><br>$\Rightarrow$ $a$ = ${1 \over 4}$ <br><br>$\Rightarrow$ ${\alpha ^2}$ = 20 <br><br>$\alpha$$\beta$ = -10 <br><br>$\Rightarrow$${\beta ^2}$ = 5 <br><br>$\therefore$ ${\alpha ^2} + {\beta ^2}$ = 20 + 5 = 25