### Video Transcript

Find algebraically the solution set
of the equation the absolute value of π₯ plus three times the absolute value of π₯
minus three equals 39.

In this question, we need to work
out the solution to two absolute value expressions multiplied together. Letβs consider the different
possible values for each expression in turn. The absolute value of π₯ plus three
will either be π₯ plus three or negative π₯ plus three. It will be π₯ plus three when π₯
plus three is greater than or equal to zero. We can simplify this by subtracting
three from both sides so that the inequality is π₯ is greater than or equal to
negative three. And we get a negative result when
π₯ plus three is less than zero or π₯ is less than negative three.

Next, for the absolute value of π₯
minus three, weβll either have π₯ minus three or negative π₯ minus three. These will occur when π₯ minus
three is greater than or equal to zero or when itβs less than zero. These inequalities can
alternatively be written as when π₯ is greater than or equal to three or π₯ is less
than three.

Now we can consider the different
possibilities of expressions that will multiply together. To begin with, we could have π₯
plus three multiplied by π₯ minus three equals 39. Or we could have π₯ plus three
multiplied by negative π₯ minus three equals 39. The third option is that we have
negative π₯ plus three multiplied by π₯ minus three equals 39. Or finally, we could have negative
π₯ plus three multiplied by negative π₯ minus three equals 39.

Before we rush into solving all
four equations, there is something we can observe. Looking at the second and third
equation, because of the commutative property of multiplication, this negative here
could equivalently be put here and the product would not change. That means that the equations two
and three would produce the same result.

And then what do we notice about
the first and final equations. Observe that this negative π₯ plus
three multiplied by a negative π₯ minus three would produce a positive. In other words, itβs the same as π₯
plus three multiplied by π₯ minus three.

So letβs have a look at solving the
first equation, and we know that it would give the same result as the fourth
equation. We could use a method such as the
FOIL method to expand the parentheses on the left-hand side, observing that the
terms plus three π₯ minus three π₯ would simplify to zero. Thus, π₯ squared minus nine equals
39. We can then add nine to both
sides. We could then take the square root
of both sides, so weβd have that π₯ is equal to the square root of 48. However, as we want to consider
both the positive and negative values of the square root, then we can use the plus
or minus symbol. As 48 can be written as 16
multiplied by three, it means that we can write this more simply as π₯ equals plus
or minus four root three.

Now letβs take a look at solving
one of the other equations, either equation two or three. Choosing negative π₯ plus three
multiplied by π₯ minus three equals 39, we can begin by expanding the
parentheses. Using our previous result that π₯
plus three multiplied by π₯ minus three would give us π₯ squared minus nine, we have
that negative π₯ squared minus nine equals 39. Multiplying both sides of this
equation by negative one would give us π₯ squared minus nine equals negative 39. Then adding nine would give us π₯
squared equals negative 30.

At this point, we may notice that
we have a problem. For as we begin to take the square
root of both sides, we see that weβre trying to take the square root of a negative
value. In this case, there would be no
real solution for π₯, so this solution would not be valid.

Before we finalize our answer,
letβs just think about whether or not the two solutions to the equation π₯ plus
three times π₯ minus three equals 39 are valid for the absolute value equation. This particular equation only
applies when π₯ is less than negative three or when π₯ is greater than or equal to
three. So if the solution for π₯ is in the
region negative three is less than or equal to π₯ is less than three, then it wonβt
be valid. As it turns out, negative four root
three is less than negative three; itβs about negative 6.93. And four root three is greater than
or equal to three; itβs about 6.93. Both of these solutions are
valid. We therefore just have one set of
values for the solution to this equation.

The solution set will be negative
four root three and four root three.