# The value of x satisfying the equation abs(abs(abs(x^2-x+4)-2)-3)=x^2+x-12 is?

Jul 30, 2018

$x = \frac{11}{2}$

#### Explanation:

Given:

$\left\mid \left\mid \left\mid {x}^{2} - x + 4 \right\mid - 2 \right\mid - 3 \right\mid = {x}^{2} + x - 12$

Note that:

${x}^{2} + x - 12 = \left(x + 4\right) \left(x - 3\right)$

which is non-negative (as required) when $x \le - 4$ or $x \ge 3$

Also note that:

${x}^{2} - x + 4 = {\left(x - \frac{1}{2}\right)}^{2} + \frac{15}{4} > 2 \text{ }$ for all $x \in \mathbb{R}$

So:

$\left\mid \left\mid {x}^{2} - x + 4 \right\mid - 2 \right\mid = \left\mid {\left(x - \frac{1}{2}\right)}^{2} + \frac{15}{4} - 2 \right\mid = {\left(x - \frac{1}{2}\right)}^{2} + \frac{7}{4}$

If $x \le - 4$ or $x \ge 3$ then ${\left(x - \frac{1}{2}\right)}^{2} \ge {\left(\frac{5}{2}\right)}^{2} = \frac{25}{4} > 3$

So when $x \le - 4$ or $x \ge 3$:

$\left\mid \left\mid \left\mid {x}^{2} - x + 4 \right\mid - 2 \right\mid - 3 \right\mid = {x}^{2} - x - 1$

and we want to solve:

${x}^{2} - x - 1 = {x}^{2} + x - 12$

Subtracting ${x}^{2} - x - 12$ from both sides, this becomes:

$11 = 2 x$

Hence:

$x = \frac{11}{2} > 3$

graph{(y-abs((x^2-x+2)-3))(y-(x^2+x-12)) = 0 [-7.7, 12.3, -6, 32]}