Question

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

• A. $1$
• B. $-1$
• C. $-6$
• D. $6$

Answer 1

Answer: (B)

$-1$

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

Let $t=x^2+x-2$

$\Rightarrow \left(t\right)(t-1)-12=0$

$\Rightarrow t^2-t-12=0$

$\Rightarrow t^2-4t+3t-12=0$

$\Rightarrow t(t-4)+3(t-4)=0$

$\Rightarrow (t-4)(t+3)=0$

$\Rightarrow t=4$ and $t=-3$

$\Rightarrow x^2+x-2=4$ and $x^2+x-2=-3$

$\Rightarrow x^2+x-6=0$ and $x^2+x-2+3=0$

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

$\Rightarrow x^2+3x-2x-6=0$ and $x^2+2\times x\times \frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1=0$

$\Rightarrow x(x+3)-2(x+3)=0$ and ${(x-\frac{1}{2})}^2+\frac{-1+4}{4}=0$

$\Rightarrow (x+3)(x-2)=0$ and ${(x-\frac{1}{2})}^2=\frac{-3}{4}$

$\Rightarrow x=-3,2$ and $x=-\frac{1}{2}\pm \sqrt{\frac{-3}{4}}$

Sum$=-3+2=-1$