# Question #2b050

Jun 24, 2015

Square the both sides

#### Explanation:

$| {x}^{2} - 3 x | = - 4 x + 6$
$\sqrt{{\left({x}^{2} - 3 x\right)}^{2}} = - 4 x + 6$
${x}^{2} {\left(x - 3\right)}^{2} = {\left(- 4 x + 6\right)}^{2}$
${x}^{2} \left({x}^{2} - 6 x + 9\right) = 16 {x}^{2} - 48 x + 36$
${x}^{4} - 6 {x}^{3} + 9 {x}^{2} - 16 {x}^{2} + 48 x - 36 = 0$
${x}^{4} - 6 {x}^{3} - 7 {x}^{2} + 48 x - 36 = 0$
As sum of all coefficients are zero, x-1 is a factor.
$\left(x - 1\right) \left({x}^{3} - 5 {x}^{2} - 12 x + 36\right) = 0$
The rest of the cubic polynomial can be factored as $6 \cdot 2 \cdot - 3$
So the roots are now $x = 1 , 2 , 6 , - 3$