Question

If $x^2-x-2$ is factor of $x^4-\lambda x^2-\mu$, then $\sqrt{({\lambda }^2-{\mu }^2)}$ equals.

• A. 1
• B. 3
• C. 5
• D. 7

Answer 1

The correct option is B

3

$\because x^2-x-2=0$

$\Rightarrow x=2,-1$

Then $x=2,-1$ are the roots of $x^4-\lambda x^2-\mu =0$

then $(2{)}^4-\lambda (2{)}^2-\mu =0$

$\Rightarrow 4\lambda +\mu =16$

and $(-4{)}^4-\lambda (-1{)}^2-\mu =0$

$\Rightarrow \lambda +\mu =1$

we get $\lambda =5$ and $\mu =-4$

$\therefore \sqrt{({\lambda }^2-{\mu }^2)}=\sqrt{25-16}=\sqrt{9}=3$