Question Video: Using Quadratic Equations to Solve Problems Mathematics • 9th Grade

The diagram shows a rectangular prism, where the area of its net is 580. Find the value of 𝑥.


Video Transcript

The diagram shows a rectangular prism, where the area of its net is 580. Find the value of 𝑥.

We’re given a rectangular prism or a cuboid, and we’re told the area of its net. Let’s think about what its net would look like. A cuboid has six faces made up of three pairs of identical faces. The front and back faces are rectangles with dimensions of 𝑥 units and three units. The two sides of the cuboid’s are also rectangles, this time with dimensions of two 𝑥 and three units. And finally, the top and base of the cuboid are another pair of rectangles, this time with dimensions 𝑥 and two 𝑥 units.

We can sketch the net of the cuboid, and it looks something like this. Now we’re told that the area of the net is 580, so we need to find an expression for the area of the net in terms of 𝑥. The orange rectangles will each have an area of three multiplied by 𝑥. That’s three 𝑥 square units. The pink rectangles will each have an area of three multiplied by two 𝑥. That’s six 𝑥 square units. And the green rectangles will each have an area of 𝑥 multiplied by two 𝑥. That’s two 𝑥 squared square units. We can now form an equation by summing each of these areas and equating to 580.

This gives two multiplied by two 𝑥 squared plus two multiplied by six 𝑥 plus two multiplied by three 𝑥 is equal to 580. Now at this stage, we can simplify our equation by dividing every term by two. And we obtain two 𝑥 squared plus six 𝑥 plus three 𝑥 equals 290. The next step in simplifying this equation is to collect the like terms on the left-hand side, which gives two 𝑥 squared plus nine 𝑥 equals 290.

The final step is to subtract 290 from each side of the equation so that all of the terms are grouped on the left-hand side. We have then two 𝑥 squared plus nine 𝑥 minus 290 is equal to zero. What we found is a quadratic equation in 𝑥. And we need to solve this equation. There are a variety of ways we could do this. We could see if the equation could be solved by factoring, we could apply the quadratic formula, or we could try completing the square.

Now, usually, if a quadratic equation can be solved by factoring, then this is the most efficient method. We’re looking for two linear factors, each an expression in 𝑥, which multiply to give the original quadratic. As the coefficient of 𝑥 squared in this equation is two, which is a prime number, we know that the first term in one set of parentheses must be two 𝑥 and the first term and the other must be 𝑥 because two 𝑥 multiplied by 𝑥 gives two 𝑥 squared.

To complete the parentheses, we’re then looking for two values which have a product equal to the constant term in the equation, which is negative 290. We can determine the possibilities for these numbers by listing the factor pairs of 290, which are one and 290, two and 145, five and 58, and 10 and 29. Remember, though, we want the product to be negative 290, which means we’ll need the numbers to have different signs; one must be positive and the other, negative.

Now whichever pair of factors we use, we need to ensure that when we distribute the parentheses, the coefficient of 𝑥 is the same as the coefficient of 𝑥 in the original equation, which was positive nine. We can use trial and error to work out which pair of factors we need. For example, suppose we chose the factor pair of 58 and five first of all. And we put positive 58 in the first set of parentheses and negative five in the second set of parentheses. We know that when we expand, we’re going to get two 𝑥 squared and negative 290. But what about the coefficient of 𝑥?

Well, when we do two 𝑥 times negative five, the contribution to the coefficient here will be negative 10. And when we do 58 times 𝑥, the contribution to the coefficient here will be 58. The coefficient of 𝑥 then will be negative 10 plus 58, which is 48. And that’s not what we’re looking for.

If, however, we choose the final factor pair and we make the 10 negative and put it in the second set of parentheses, then we know again that we’re going to get two 𝑥 squared and negative 290 when we distribute the parentheses. And for the coefficient of 𝑥, we’ll have 29 multiplied by one and two multiplied by negative 10. That’s 29 minus 20, which is equal to nine. So we have the correct coefficient of 𝑥. This tells us that this combination of factors and signs is the correct factorization. We can, of course, confirm this by redistributing the parentheses if we wish.

Next, we recall that if a product is equal to zero, then at least one of the individual factors must themselves be equal to zero. So we have either two 𝑥 plus 29 equals zero or 𝑥 minus 10 equals zero. These are linear equations in 𝑥, which we can solve. To solve to the first equation, we subtract 29 from each side and then divide by two, giving 𝑥 equals negative 29 over two. To solve the second equation, we can do this in one step by adding 10 to each side to give 𝑥 equals 10.

We’ve found two solutions to this quadratic equation then. But are they both possible values of 𝑥 in this context? If we look back at the diagram, we can see that 𝑥 represents the length of one of the sides in this cuboid. And so 𝑥 must have a positive value. This tells us that whilst negative 29 over two is a valid solution to this quadratic equation, it’s not the value of 𝑥 we’re looking for in the context of this problem. By forming and solving a quadratic equation, which we did by factoring the equation, we found that the value of 𝑥 is 10.

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