Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $6$

Answer 1

The correct option is A $1$

Let $x^2+x-2=t$
$\Rightarrow t(t-1)=12\Rightarrow t=4,-3$

$\therefore x^2+x-2=4$ or $x^2+x-2=-3$
$\Rightarrow x^2+x-6=0$ or $x^2+x+1=0$
$x^2+x+1=0$ has non-real roots.
$\therefore$ Product of non-real roots $=1$