Question

Solve |p + 2| = 10Solve |p + 2| = 10 {-12} {-8, 8} {-12, 8}

Answer 1

By definition, we have

$|p+2| = \begin{cases} p+2 &\text{ if } p+2 \geq 0 \\-p-2 &\text{ if } p+2 < 0 \end{cases}$

So, we have to solve two different equations, depending of the possible range for the variable. We have to remember about these ranges when we decide to accept or discard the solutions:

Suppose that $p+2\geq 0 \iff p \geq -2$

In this case, the absolute value doesn't do anything: the equation is

$p+2 = 10 \iff p = 10-2 = 8$

We are supposing $p \geq -2$, so we can accept this solution.

Now, suppose that $p+2 < 0 \iff p < -2$. Now the sign of the expression is flipped by the absolute value, and the equation becomes

$-p-2 = 10 \iff -p = 12 \iff p = -12$

Again, the solution is coherent with the assumption, so we can accept this value as well.