### Quadratic Equation Question 7

#### Question 7 - 30 January - Shift 2

If the value of real number $a>0$ for which $x^{2}-5 a x$ $+1=0$ and $x^{2}-a x-5=0$ have a common real roots is $\frac{3}{\sqrt{2 \beta}}$ then $\beta$ is equal to______

## Show Answer

#### Answer: 13

#### Solution:

#### Formula: Common Roots (ii)

Let $\alpha$ be the common roots

$ \therefore \alpha^2+\alpha+a=0 $ and $ \alpha^2+a \alpha+1=0 $

$ \frac{\alpha^2}{1-a^2}=\frac{\alpha}{a-1}=\frac{1}{a-1}[a \neq 1] $

Eliminating $\alpha$, we get

$ \begin{aligned} & (a-1)^2=\left(1-a^2\right)(a-1) \\ & \Rightarrow(a-1)=1-a^2 \\ & \Rightarrow a^2+a-2=0 \\ & \Rightarrow a^2+2 a-a-2=0 \\ & \Rightarrow a(a+2)-1(a+2)=0 \\ & \Rightarrow(a+2)(a-1)=0 \\ & \Rightarrow a=-2 \quad[\because a \neq 1] \end{aligned} $