 # Prove the following Question:

Let $a, b \in \mathrm{R}, a \neq 0$ be such that the equation, $a x^{2}-2 b x+5=0$ has a repeated root $\alpha$, which is also a root of the equation, $x^{2}-2 b x-10=0$. If $\beta$ is the other root of this equation, then $\alpha^{2}+\beta^{2}$ is equal to :

1. (1) 25

2. (2) 26

3. (3) 28

4. (4) 24

Correct Option: 1

Solution:

$a x^{2}-2 b x+5=0$

If $\alpha$ and $\alpha$ are roots of equations, then sum of roots

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

and product of roots $=\alpha^{2}=\frac{5}{a} \quad \Rightarrow \quad \frac{b^{2}}{a^{2}}=\frac{5}{a}$

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

For $x^{2}-2 b x-10=0$

$\alpha+\beta=2 b$ . ...(ii)

and $\alpha \beta=-10$ ...(iii)

$\alpha=\frac{b}{a}$ is also root of $x^{2}-2 b x-10=0$

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

By eqn. (i) $\Rightarrow 5 a-10 a^{2}-10 a^{2}=0$

$\Rightarrow 20 a^{2}=5 a$

$\Rightarrow \quad a=\frac{1}{4}$ and $b^{2}=\frac{5}{4}$

$\alpha^{2}=20$ and $\beta^{2}=5$

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