Question

If the roots of the given equation $2x^2+3(\lambda -2)x+\lambda +4=0$ be equal in magnitude but opposite in sign, then value of $\lambda$ is

• A. $1$
• B. $2$
• C. $3$
• D. $\frac{2}{3}$

Answer 1

Answer: (B)

$2$

Given equation is $2x^2+3(\lambda -2)x+\lambda +4=0$

We know that equation $a{x}^2+bx+c=0$

Then sum of roots $=\frac{-b}{a}$

Let $\alpha$, $\beta$ be the roots of the given equation

$\therefore \alpha +\beta =-\frac{3(\lambda -2)}{2}$

But it is given that
$\alpha =-\beta$
Therefore, $-\beta +\beta =-\frac{3(\lambda -2)}{2}$

$0=-\frac{3(\lambda -2)}{2}$

$0\times 2=-3(\lambda -2)$
$\frac{0}{-3}=(\lambda -2)$
$0=(\lambda -2)$
$\lambda -2=0$
Hence, $\therefore \lambda =2$