# How do you solve abs(2v)=abs(-16-3v)?

Jun 2, 2015

You can solve such problems by splitting into cases where the sign of the enclosed expressions changes, but on this occasion I will square both sides instead to get:

$4 {v}^{2} = 256 + 96 v + 9 {v}^{2}$

Then subtract $4 {v}^{2}$ from both sides to get:

$5 {v}^{2} + 96 v + 256 = 0$

Next use a modified AC Method to find factors:

$A = 5$, $B = 96$, $C = 256$

Look for a factorization of $A C = 5 \cdot 256$ into a pair of factors whose sum is $B = 96$.

With a little reasoning, I quickly found the pair $B 1 = 16$, $B 2 = 80$

Then for each combination of $A , B 1$ and $A , B 2$ divide by their HCF (highest common factor) to give a pair of coefficients of a factor:

$\left(A , B 1\right) = \left(5 , 16\right)$ HCF $1$ $\to \left(5 , 16\right) \to \left(5 v + 16\right)$

$\left(A , B 2\right) = \left(5 , 80\right)$ HCF $5$ $\to \left(1 , 16\right) \to \left(v + 16\right)$

So $5 {v}^{2} + 96 v + 256 = \left(5 v + 16\right) \left(v + 16\right)$

This has zeros for $v = - 16$ and $v = - \frac{16}{5}$

Check:

$v = - 16$ :

$\left\mid 2 v \right\mid = 32 = \left\mid - 16 + 48 \right\mid = \left\mid - 16 - 3 v \right\mid$

$v = - \frac{16}{5}$ :

$\left\mid 2 v \right\mid = \frac{32}{5} = \left\mid - 16 + 16 - \frac{32}{5} \right\mid$

$= \left\mid - 16 + \frac{80 - 32}{5} \right\mid = \left\mid - 16 + \frac{48}{5} \right\mid$

$= \left\mid - 16 + \frac{3 \cdot 16}{5} \right\mid = \left\mid - 16 - 3 v \right\mid$