Question

Find the value of $m$ for which the equations $x^2+mx+1=0\&(b-c)x^2+(c-a)x+(a-b)=0$ have both roots common.

• A. $-2$
• B. $1$
• C. $2$
• D. $0$

Answer 1

The correct option is A $-2$

Given: $x^2+mx+1=0\cdots \left(i\right)$
$\&(b-c)x^2+(c-a)x+(a-b)=0\cdots \left(ii\right)$
Sum of coefficients of $\left(ii\right)$ is zero
$\Rightarrow x=1$ is one the roots of eq. $\left(ii\right)$
Since, the equations have both roots same.
$\therefore x=1$ will be the root of equation $\left(i\right)$ too.
$\Rightarrow (1{)}^2+m.1+1=0$
$\Rightarrow 1+m+1=0$
$\Rightarrow m=-2$