The sum of the non-real roots of $\left(p^2+p-3\right)\left(p^2+p-2\right)-12=0$ is
The sum of the non-real roots of $\left(p^2+p-3\right)\left(p^2+p-2\right)-12=0$ is
- 1
- -1
- 6
- -6
Solution
Given equation is
$
\left(P^2+P-3\right)\left(P^2+P-2\right)-12=0
$
Let $y=P^2+P-2$
$
\begin{array}{lr}
\therefore & y(y-1)-12=0 \\
\Rightarrow & y^2-y-12=0 \\
\Rightarrow & (y-4)(y+3)=0 \\
\Rightarrow & y-4=0 \text { as } y+3=0 \\
\Rightarrow & P^2+P-6=0 \text { as } P^2+P+1=0
\end{array}
$
For imaginary (non-real roots)
$
P^2+P+1=0
$
$\therefore$ Sum of non-real roots $=-1$
Asked in: AP EAMCET 2018 (24 Apr Shift 1)
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