## Distance From Zero

Find the product of all real values of $p$ that satisfy the equation $6|p-6|=|p+6|$.
Source: NCTM Mathematics Teacher, January 2006

SOLUTION
One way to interpret the absolute values $|p-6|$ and $|p+6|$ is to think of them as the distances from point $P$ to point $B$ and from point $P$ to point $A$ on the number line

$|p-6|=PB$
$|p+6|=|p-(\textrm{-}6)|=PA$
We want to find the points $P$ such that $PA=6PB$.
Case 1: $P$ is between $A$ and $B$

If we divide segment $\overline{AB}$ into $7$ equal parts by drawing a series of parallel lines and place $P$ as shown in the above figure, then clearly $PA=6PB$.
$p=\textrm{-}6+6(12/7)=30/7$
Case 2: $P$ is not between $A$ and $B$

If we divide segment $\overline{AB}$ into $5$ equal parts and place $P$ by the same equal part from B, then $PA=6PB$.
$p=\textrm{-}6+6(12/5)=42/5$
Product of all $p$ values
$30/7\times 42/5=36$

Answer: $36$

Alternative solution 1

If we plot the two functions $y_1=|x+6|$ and $y_2=6|x-6|$, the graphs intersect at two points $(4.3,10.3)$ and $(8.4,14.4)$.
Multiplying the two $x$ values yields
$4.3\times 8.4=36.12$

Alternative solution 2
If $0 and $p+6>0$.
$6|p-6|=|p+6|$
$6(\textrm{-}p+6)=p+6$
$\textrm{-}6p+36=p+6$
$\textrm{-}7p=\textrm{-}30$
$p=30/7$
If $p>6, p-6>0$ and $p+6>0$.
$6|p-6|=|p+6|$
$6(p-6)=p+6$
$6p-36=p+6$
$5p=42$
$p=42/5$
$30/7\times 42/5=36$