Question

The value of $x$ , which satisfies the equation $\left(x - 1\right)\left|x^{2} - 4 x + 3\right|+2x^{2}+3x-5=0$ , is

Answer 1

$\left(x - 1\right)\left|x^{2} - 4 x + 3\right|+2x^{2}+3x-5=0$
Case I: When $x^{2}-4x+3\geq 0$ , then $x\in \left(\right.-\infty ,1\left]\right.\cup\left[\right.3,\infty \left.\right)$
$\Rightarrow \left(x - 1\right)\left(x^{2} - 3 x - x + 3\right)+2x^{2}+5x-2x-5=0$
$\Rightarrow \left(x - 1\right)\left(x^{2} - 3 x - x + 3 + 2 x + 5\right)=0$
$\left(x - 1\right)\left(x^{2} - 2 x + 8\right)=0$ (the value of quadratic cannot be zero since its $D < 0$ )
$\therefore x-1=0$
$\Rightarrow x=1$
Case II: When $\text{x}^{2} - 4 \text{x} + 3 < 0$ , then $x\in \left(1 , 3\right)$
$\left(x - 1\right)\left(- x^{2} + 4 x - 3 + 2 x + 5\right)=0$
$x-1≠0$
$\therefore -x^{2}+6x+2=0$
$\Rightarrow x^{2}-6x-2=0$
$\Rightarrow x=3\pm\sqrt{1 1}$
Both values will not satisfy
So, $x=1$