### Video Transcript

If the area of the given figure is 138 square centimeters, what is the value of 𝑥?

The diagram shows a composite shape. We can divide it into two rectangles either using a horizontal or vertical line. If we divide the shape here, then we create two rectangles. We can find expressions for the area of each of these rectangles by multiplying their dimensions together. The rectangle at the bottom is the more straightforward one. Its dimensions are 15 centimeters and 𝑥 centimeters. So its area is 15𝑥 square centimeters. For the rectangle at the top, one of its dimensions is 𝑥 centimeters, but what about the other?

Well, if the total height of this figure is 14 centimeters and the height of the bottom part is 𝑥 centimeters, then the height of the top rectangle will be 14 minus 𝑥 centimeters. And so the area of the top rectangle found by multiplying its dimensions together will be 𝑥 multiplied by 14 minus 𝑥 square centimeters.

Now, we’re told what the area of this figure is. It’s 138 square centimeters. And so we can form an equation by adding our expressions for the two areas together. We have 𝑥 multiplied by 14 minus 𝑥 plus 15𝑥 is equal to 138. To find the value of 𝑥, we’re going to need to solve this equation. Let’s begin by distributing the set of parentheses on the left-hand side. This gives 14𝑥 minus 𝑥 squared plus 15𝑥 equals 138. Next, we’re going to group the like terms. We have 14𝑥 and positive 15𝑥. So our equation is negative 𝑥 squared plus 29𝑥 equals 138. Next, we can collect all of the terms on the same side of the equation by subtracting 138 from each side, giving negative 𝑥 squared plus 29𝑥 minus 138 equals zero.

Now, this is a quadratic equation in 𝑥. But it’s a little tricky because the leading coefficient — in this case, that’s the coefficient of 𝑥 squared — is negative. However, we can change this. If we multiply the entire equation through by negative one, then the coefficient of 𝑥 squared will be positive. Doing so will change the signs of every term in the equation. So we now have 𝑥 squared minus 29𝑥 plus 138 is equal to zero.

Now, there are a number of methods that we can use to solve a quadratic equation. We can try factoring, we can try the quadratic formula, or we can complete the square. If a quadratic equation can be solved by factoring, then that’s usually the most straightforward method. So let’s try that first.

We’re looking for two linear factors that multiply together to give this quadratic. As the coefficient of 𝑥 squared in our equation is one, the first term in each set of parentheses will be 𝑥, because 𝑥 multiplied by 𝑥 gives 𝑥 squared. We’re then looking for two values to complete our parentheses which have a specific set of properties. First, the sum of these two numbers must be equal to the coefficient of 𝑥. That’s negative 29. And second, their product must be equal to the constant term, which is positive 138.

Now, it’s a little tricky to think of these numbers off the top of our head. So let’s start by listing the factor pairs of 138. They are one and 138, two and 69, three and 46, and six and 23. We notice that the product of the two numbers must be positive. So they must have the same sign; they’re either both positive or both negative. We then notice that the sum of the two numbers must be negative. So the two numbers must themselves both be negative. We’re then looking for the factor pair which also has a sum of negative 29. And we see it is the final factor pair. Negative six plus negative 23 is negative 29. So these are the two numbers we need to complete our parentheses.

We find then that the quadratic factors as 𝑥 minus six multiplied by 𝑥 minus 23. And we can check this by redistributing the parentheses using the FOIL method if we wish.

Next, we recall that, for a product to be equal to zero, one of the individual factors must be equal to zero. So either 𝑥 minus six equals zero or 𝑥 minus 23 is equal to zero. Each of these equations can be solved in one step. To solve the first equation, we add six to each side, giving 𝑥 equals six. And to solve the second, we add 23 to each side, giving 𝑥 equals 23. There are, therefore, two solutions to this quadratic equation, but are they both possible values for 𝑥 in this context?

Looking back at the figure, we see that one of the side lengths in the diagram is represented by the expression 14 minus 𝑥. This tells us that the value of 𝑥 must be less than 14, because we can’t have a negative value for a length. So whilst 𝑥 equals 23 is a valid solution to this quadratic equation, it isn’t the value we’re looking for in the context of this problem. And our answer must be the other value of 𝑥. So our answer is six.

We can check this though by evaluating the expressions for each of the lengths in the figure and confirming that the area is indeed equal to 138 square centimeters. If 𝑥 is equal to six, then the dimensions of the first rectangle are six centimeters and eight centimeters. So its area is equal to 48 square centimeters. The dimensions of the second rectangle are 15 centimeters and six centimeters. So the area of this rectangle is 90 square centimeters. 48 plus 90 is equal to 138. So this confirms that our answer is correct. The value of 𝑥 found by forming and then solving a quadratic equation is six.