Question

Find the sum of the non-real roots of the equation $(x^2+x-2)(x^2+x-3)=12$

Answer 1

We have,

$(x^2+x-2)(x^2+x-3)=12$

Let, $(x^2+x-2)=y$

Then,

$y(y-1)=12$

$y^2-y=12$

$y^2-y-12=0$

$y^2-(4-3)y-12=0$

$y^2-4y+3y-12=0$

$y(y-4)+3(y-4)=0$

$(y-4)(y+3)=0$

If,$y-4=0$ then,$y=4$

If,$y+3=0$ then,$y=-3$

Put the value of $y=(x^2+x-2)$

$(x^2+x-2)=4$

$x^2+x-6=0$

$x^2+(3-2)x-6=0$

$x^2+3x-2x-6=0$

$x(x+3)-2(x+3)=0$

$(x+3)(x-2)=0$

$x=2,-3$

If, $y=-3$

$x^2+x-2=-3$

$x^2+x-1=0$

Then, sum of non real values$=-1$

Hence, this is the answer.