Q. Solution set of $\left|x^2-5 x+7\right|+\left|x^2-5 x-14\right|=21$ is
Complex Numbers and Quadratic Equations
Solution:
$\left|x^2-5 x+7\right|+\left|4+5 x-x^2\right|=21$
$|x|+|y|=|x+y| $
$x y \geq 0$
$\left(x^2-5 x+7\right)\left(14+5 x-x^2\right) \geq 0$
$\left(x^2-5 x+7\right)\left(x^2-5 x-14\right) \leq 0$
$x^2-5 x=t $
$(t+7)(t-14) \leq 0 $
$ t \in[-7,14] $
$-7 \leq x^2-5 x \leq 14$
