Question

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

Answer 1

Answer: 1

$\left(x-1\right)\left|x^2-4x+3\right|+2x^2+3x-5=0$
Case I: When $x^2-4x+3\ge 0$ , then $x\in (-\infty ,1]\cup [3,\infty )$
$\Rightarrow \left(x-1\right)\left(x^2-3x-x+3\right)+2x^2+5x-2x-5=0$
$\Rightarrow \left(x-1\right)\left(x^2-3x-x+3+2x+5\right)=0$
$\left(x-1\right)\left(x^2-2x+8\right)=0$ (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 $x\in \left(1,3\right)$
$\left(x-1\right)\left(-x^2+4x-3+2x+5\right)=0$
$x-1\ne 0$
$\therefore -x^2+6x+2=0$
$\Rightarrow x^2-6x-2=0$
$\Rightarrow x=3\pm \sqrt{11}$
Both values will not satisfy
So, $x=1$