Let a,b∈ R,a≠ 0 be such that the equation, ax^2-2bx+5=0 has a repeated root α , which is also a root of the equation, x^2-2bx-10=0. If β is the other…

Given, $a{x}^{2}-2bx+5=0$ax2−2bx+5=0 has repeated root $\alpha$α.

$\therefore 2\alpha =\frac{2b}{a}\Rightarrow \alpha =\frac{b}{a}$∴2α=a2b​⇒α=ab​and ${\alpha }^{2}=\frac{5}{a}\Rightarrow \frac{{b}^{2}}{{a}^{2}}=\frac{5}{a}$α2=a5​⇒a2b2​=a5​

$\Rightarrow {b}^{2}=5a...(i)$⇒b2=5a...(i) $(a\neq 0)$(a=0)

$$\alpha +\beta =2b...(\mathrm{ii})$$α+β=2b...(ii)

and $\alpha \beta =-10...(\mathrm{iii})$αβ=−10...(iii)

$\alpha =\frac{b}{a}$α=ab​ is also root of ${x}^{2}-2bx-10=0$x2−2bx−10=0

$$\Rightarrow {b}^{2}-2a{b}^{2}-10{a}^{2}=0$$⇒b2−2ab2−10a2=0

by $(i)$(i)$\Rightarrow 5a-10{a}^{2}-10{a}^{2}=0$⇒5a−10a2−10a2=0

$$\Rightarrow 20{a}^{2}=5a$$⇒20a2=5a

$\Rightarrow a=\frac{1}{4}$⇒a=41​ and ${b}^{2}=\frac{5}{4}$b2=45​

Now ${\alpha }^{2}+{\beta }^{2}={(\alpha +\beta )}^{2}-2\alpha \beta$α2+β2=(α+β)2−2αβ

$$=5+20$$=5+20

$$=25$$=25