Question

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

• A. $1$
• B. $-1$
• C. $6$
• D. $-6$

Answer 1

The correct option is B $-1$

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

Let $x^2+x=a\to 1$

$(a-2)(a-3)=12$

$a^2-5a+6=12$

$a^2-5a-6=0$

$(a+1)(a-6)=0$

$\therefore a=-1$ or $a=6$

If $a=6$

$1\Rightarrow x^2+x-6=0$

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

$\therefore x=-3$ or $2$

If $a=-1$

$1\Rightarrow x^2+x+1=0$

$△=1-4=-3<0$

$\therefore$Roots are complex i.e., non real

$\therefore$Sum of non-real roots$=-1$