Question

Sum of the non-real roots of $(x^2+x-2)(x^2+x-3)=12$ is

• A. $-1$
• B. $1$
• C. $-2$
• D. $2$

Answer 1

The correct option is A $-1$

$(x^2+x-2)(x^2+x-3)=12$
Put $x^2+x=y$
$(y-2)(y-3)=12$
$\Rightarrow y^2-5y-6=0$
$\Rightarrow (y-6)(y+1)=0$
$\Rightarrow y=6,-1$

When $y=6$,
$x^2+x-6=0$
$\Rightarrow (x+3)(x-2)=0$
$\Rightarrow x=-3,2$

When $y=-1$,
$x^2+x+1=0$ which has non-real roots.
Sum of roots is $-1$.