Let a, b \in R, a e 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 :

Quadratic Equation and Inequalities PYQ — JEE Main 2020 · 9 Jan · Shift 2 · Q35 with Solution

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Correct answer: D

Since the quadratic $ax^2-2bx+5=0$ has a repeated root $\alpha$ , its discriminant must be zero.

So, $(-2b)^2-4(a)(5)=0$ $4b^2-20a=0$ $b^2=5a$

Also, for a quadratic with repeated root $\alpha$ , $\alpha=\frac{-(-2b)}{2a}=\frac{2b}{2a}=\frac{b}{a}$

We are also given that $\alpha$ is a root of $x^2-2bx-10=0$ So, $\alpha^2-2b\alpha-10=0$

Substitute $\alpha=\frac{b}{a}$ : $\left(\frac{b}{a}\right)^2-2b\left(\frac{b}{a}\right)-10=0$

Multiply by $a^2$ : $b^2-2ab^2-10a^2=0$

Now use $b^2=5a$ : $5a-2a(5a)-10a^2=0$ $5a-10a^2-10a^2=0$ $5a-20a^2=0$ $5a(1-4a)=0$

Given $a\ne 0$ , $a=\frac14$ Then $b^2=5a=\frac54$

\qquad \alpha\beta=-10$$ We need $$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$$ $$=(2b)^2-2(-10)$$ $$=4b^2+20$$ Using $b^2=\frac54$, $$\alpha^2+\beta^2=4\cdot \frac54+20=5+20=25$$ 4. Therefore the correct option is $$\boxed{25}$$ which is option $\boxed{D}$.