If the value of real number $a0$ for which $x^{2}-5 a x+1=0$ and $x^{2}-a x-5=0$ have a common real

Question

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 root is $\frac{3}{\sqrt{2 \beta}}$ then $\beta$ is equal to

Professor PC Thomas Classes · Answer

(13) : $x^{2}-5 a x+1=0$ Here $a_{1}=1, b_{1}=-5 a, c_{1}=1$ $$\begin{align*} & x^{2}-a x-5=0  	ag{ii}\ & a_{2}=1, b_{2}=-a, c_{2}=-5 \end{align*}$$ $\because \quad$ Equation (i) and (ii) have one common real root, so $$ \begin{aligned} & \quad[1 	imes 1-(-5)(1)]^{2}=[(-5 a)(-5)-(-a) \ & (1)](-a+5 a) \ & \Rightarrow \quad 36=(25 a+a)(4 a) \Rightarrow 36=26 a 	imes 4 a \ & 	herefore \quad a^{2}=\frac{36}{26 	imes 4} \Rightarrow a= \pm \frac{3}{\sqrt{26}} \end{aligned} $$ Since the value of root is given $\frac{3}{\sqrt{2 \beta}}$.\ So, $\beta=13$\ So, $\beta=13$\