# Question #70e44

Apr 16, 2017

$x \in \left\{+ 5 , + 2\right\}$

#### Explanation:

Case $1$: the absolute value is positive

$3 m = {m}^{2} - 10$

$0 = {m}^{2} - 3 m - 10$

$0 = \left(m - 5\right) \left(m + 2\right)$

$m = 5 \mathmr{and} - 2$

Case $2$: the absolute value is negative

$3 m = - \left({m}^{2} - 10\right)$

$3 m = - {m}^{2} + 10$

${m}^{2} + 3 m - 10 = 0$

$\left(m + 5\right) \left(m - 2\right) = 0$

$m = - 5 \mathmr{and} 2$

However, the negative solutions are extraneous because the left hand side would be negative while the right hand side would be positive (by the definition of $| a |$).

Hopefully this helps!