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 \_\_\_\_\_\_\_\_\_\_\_.

Answer: 13

Accepted Answer

${x}^{2}-5\alpha x+1=0$ ..... (1)

${x}^{2}-\alpha x-5=0$ ...... (2)

have a common root.

Subtracting (1) with (2) we'll get $x=\frac{6}{4\alpha }$

Substituting in (1)

$\frac{36}{16{\alpha }^{2}}-\frac{30}{4}+1=0$

$\Rightarrow {\alpha }^{2}=\frac{9}{26}$

$\alpha =\frac{3}{\sqrt{2	imes 13}}$

$	herefore $ $\beta =13$

Solutions · 1