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
image
|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
image
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
image
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
image
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<p<6, p-6<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

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About mvtrinh

Retired high school math teacher.
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