Question Video: Solving Quadratic Equations Involving Absolute Values | Nagwa Question Video: Solving Quadratic Equations Involving Absolute Values | Nagwa

# Question Video: Solving Quadratic Equations Involving Absolute Values Mathematics • Second Year of Secondary School

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Find algebraically the solution set of the equation |π₯ + 3||π₯ β 3| = 39.

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### 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. Letβs 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, weβll 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 itβs 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, itβs the same as π₯ plus three multiplied by π₯ minus three.

So letβs 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 weβd 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 letβs 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 weβre 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, letβs 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 wonβt be valid. As it turns out, negative four root three is less than negative three; itβs about negative 6.93. And four root three is greater than or equal to three; itβs 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.

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