Question Video: Using Quadratic Equations to Solve Problems Mathematics

Video Transcript

The diagram shows a trapezoid and a rectangle. Part (a) Write an expression for the area of the rectangle. Part (b) Write an expression for the area of the trapezoid. Part (c) If the trapezoid and the rectangle have the same area, find the value of 𝑥 using a suitable equation.

Let’s begin with part (a) then. We need to find an expression for the area of the rectangle. We know that the area of a rectangle is found using the formula area equals length multiplied by width. We multiply the two dimensions of the rectangle together. In this problem, the dimensions of the rectangle are given by the expressions two 𝑥 plus one and 𝑥 minus nine. We can multiply in either order. So an expression for the area of the rectangle is two 𝑥 plus one multiplied by 𝑥 minus nine. This expression is unsimplified at this point. We’ll leave it in its factored form.

Next, let’s consider how we find the area of a trapezoid or trapezium. In general, the area of a trapezoid is given by its perpendicular height, ℎ, multiplied by half the sum of its parallel sides. That’s 𝑎 plus 𝑏 over two, where 𝑎 and 𝑏 are the lengths of the two parallel sides. For this trapezoid then, we have 𝑥 minus seven for the perpendicular height multiplied by 𝑥 plus 𝑥 plus six for the sum of the parallel sides over two. 𝑥 plus 𝑥 plus six is of course two 𝑥 plus six. And then we can simplify by canceling a factor of two throughout the second expression, which gives a simplified expression of 𝑥 minus seven multiplied by 𝑥 plus three. So we’ve answered the first two parts of the question.

In part (c), we’re told that the trapezoid and the rectangle have the same area. And we’re asked to find the value of this unknown 𝑥 by forming and solving a suitable equation. If the two areas are the same, then we can form an equation by taking the expressions we have for the area of the rectangle and the area of the trapezoid and setting them equal to one another. This gives the equation two 𝑥 plus one multiplied by 𝑥 minus nine equals 𝑥 minus seven multiplied by 𝑥 plus three.

We can then simplify this equation by distributing each set of parentheses. On the left-hand side, we have two 𝑥 squared minus 18𝑥 plus 𝑥 minus nine, and on the right-hand side 𝑥 squared plus three 𝑥 minus seven 𝑥 minus 21. We can then simplify on each side by grouping the like terms in the center of each expansion, giving two 𝑥 squared minus 17𝑥 minus nine is equal to 𝑥 squared minus four 𝑥 minus 21. Next, we want to group all of the terms on the same side of the equation. We’ll group the terms on the left-hand side because the coefficient of 𝑥 squared here, which is two, is greater than the coefficient of 𝑥 squared on the other side.

First, we can subtract 𝑥 squared from each side of the equation. Next, we add four 𝑥 to each side. And finally, we add 21 to each side of the equation. This will eliminate all of the terms on the right-hand side. And we’re left with 𝑥 squared minus 13𝑥 plus 12 is equal to zero.

This is a quadratic equation in our unknown variable 𝑥. Let’s see if we can solve this equation by factoring. We’re looking for two linear expressions in 𝑥 which multiply to give the original quadratic. As the coefficient of 𝑥 squared is one, we know that 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 numbers to complete these parentheses which have a specific set of properties. Firstly, their sum needs to be the coefficient of 𝑥. That’s negative 13. And secondly, their product needs to be the constant term, which is positive 12.

With a little bit of thought and perhaps by listing the factor pairs of 12, we see that the two numbers which have a sum of negative 13 and a product of 12 are negative 12 and negative one. They’re both negative so when we multiply we get a positive, 12. And when we add negative 12 and negative one, we get negative 13. Our quadratic therefore factors as 𝑥 minus 12 multiplied by 𝑥 minus one is equal to zero.

We can check this by redistributing the parentheses, perhaps using the FOIL method if we wish. Next, we recall that if the product of two values or expressions is equal to zero, then one of those expressions themselves must be equal to zero. Setting each factor in turn equal to zero leads to the equations 𝑥 minus 12 equals zero or 𝑥 minus one equals zero. Each of these equations can be solved in one step.

To solve the first equation, we add 12 to each side, giving 𝑥 is equal to 12. And to solve the second, we add one to each side, giving 𝑥 is equal to one. There are therefore two solutions to this quadratic equation. But are they both valid values for 𝑥? If we return to the diagram, we see that the height of the trapezoid is given by 𝑥 minus seven. And the width of the rectangle is given by 𝑥 minus nine. In order for these both to be positive, which they must be as they’re both lengths, we require the value of 𝑥 to be greater than nine. So whilst 𝑥 equals one is a valid solution to this quadratic equation, it isn’t a possible value for 𝑥 in the diagram. The value of 𝑥 we need then is 12.

If we want, we can check this answer by substituting 𝑥 equals 12 into the expressions for the two areas and confirming that they are indeed equal. In both cases, we get an answer of 75. So our value of 𝑥 is correct.

Our answers to the three parts of the problem then are an expression for the area of the rectangle is two 𝑥 plus one multiplied by 𝑥 minus nine. An expression for the area of the trapezoid is 𝑥 minus seven multiplied by 𝑥 plus three. And the value of 𝑥 if the trapezoid and the rectangle have the same area is 12.