Quadratic equation mixed asked by ربيع عامود in the نادي الرياضيات

Exercise:

Solve in $\mathbb{R}$ , ${{\left( {{x}^{2}}-5x+4 \right)}^{2}}-3\left( {{x}^{2}}-5x+4 \right)+1=x$

Solution: Let $u\left( x \right)={{x}^{2}}-5x+4$

So $u\left( x \right)={{x}^{2}}-4x-x+4=x\left( x-1 \right)-4\left( x-1 \right)=\left( x-1 \right)\left( x-4 \right)$

Thus ${{\left( x-1 \right)}^{2}}{{\left( x-4 \right)}^{2}}-3\left( x-4 \right)\left( x-1 \right)-\left( x-1 \right)=0$

\(\Leftrightarrow \left( x-1 \right)\left[ \left( x-1 \right){{\left( x-4 \right)}^{2}}-3\left( x-4 \right)-1 \right]=0\)

$\Leftrightarrow x=1\,\,\,\,or\,\,\,\left( x-1 \right){{\left( x-4 \right)}^{2}}-3\left( x-4 \right)-1=0$

$\Leftrightarrow x{{\left( x-4 \right)}^{2}}-{{\left( x-4 \right)}^{2}}-3\left( x-4 \right)-1=0\Leftrightarrow \left( x-4 \right)\left( x\left( x-4 \right)-3 \right)-{{\left( x-4 \right)}^{2}}-1=0$

$\Leftrightarrow \left( x-4 \right)\left( {{x}^{2}}-4x-3 \right)-{{\left( x-4 \right)}^{2}}-1=0\Leftrightarrow \left( x-4 \right)\left[ {{x}^{2}}-4x-3-x+4 \right]-1=0$

$\Leftrightarrow \left( x-4 \right)\left( {{x}^{2}}-5x+1 \right)-1=0\Leftrightarrow \left( x-4 \right)\left( x\left( x-5 \right)+1 \right)-1=0$

 $\Leftrightarrow x\left( x-5 \right)\left( x-4 \right)+x-4-1=0\Leftrightarrow \left( x-5 \right)\left( x\left( x-4 \right)+1 \right)=0$

So $x=5\,\,\,or\,\,\,{{x}^{2}}-4x+4-4+1=0\Leftrightarrow {{\left( x-2 \right)}^{2}}-3=0\Leftrightarrow x=2\pm \sqrt{3}$