Question

Find the value of $m$, if $x^2+mx+1=0$ and $(b-c)x^2+(c-a)x+(a-b)=0$ have both the roots common.

• A. $2$
• B. $1$
• C. $-2$
• D. $0$

Answer 1

The correct option is C $-2$

Given:
$x^2+mx+1=0$ ...(i) and
$(b-c)x^2+(c-a)x+(a-b)=0...\left(ii\right)$

Sum of coefficients of (ii) is zero,
$\Rightarrow x=1$ is one the roots of eq.(ii)
Since, the equations have both roots same.
$\therefore x=1$ will be the root of equation (i) also
$\Rightarrow (1{)}^2+m.1+1=0$
$\Rightarrow 1+m+1=0$
$\Rightarrow m=-2$