QJEEM20Madeec: JEE Main 2020 Mathematics - Class 11 JEEM20Madeec Solution: Let a, b R, a 0 be such that the equation, ax 2 – 2bx + 5 = 0…

Roots of equation ax² – 2bx + 5 = 0 are $\alpha$ , $\alpha$ .

$\therefore$ $\alpha$ + $\alpha$ = ${{2b} \over a}$

$\Rightarrow$ 2 $\alpha$ = ${{2b} \over a}$

$\Rightarrow$ $\alpha$ = ${{b} \over a}$ ....(1)

and $\alpha$ ² = ${5 \over a}$ ....(2)

From (1) and (2), we get

${{{b^2}} \over {{a^2}}} = {5 \over a}$

$\Rightarrow$ b² = 5a

$\alpha$ , $\beta$ are the roots of equation x² – 2bx – 10 = 0

$\therefore$ $\alpha$ + $\beta$ = 2b

and $\alpha$ $\beta$ = -10

As $\alpha$ is a root of the equation x² – 2bx – 10 = 0.

$\therefore$ $\alpha$ ² - 2b $\alpha$ - 10 = 0

$\Rightarrow$ ${{{b^2}} \over {{a^2}}} - {{2{b^2}} \over a} - 10 = 0$

$\Rightarrow$ ${{5a} \over {{a^2}}} - {{10a} \over a} - 10 = 0$

$\Rightarrow$ ${5 \over a} - 10 - 10 = 0$

$\Rightarrow$ $a$ = ${1 \over 4}$

$\Rightarrow$ ${\alpha ^2}$ = 20

$\alpha$ $\beta$ = -10

$\Rightarrow$ ${\beta ^2}$ = 5

$\therefore$ ${\alpha ^2} + {\beta ^2}$ = 20 + 5 = 25