Question Video: Forming and Solving Quadratic Equations Mathematics

Video Transcript

Determine the positive number whose square exceeds twice its value by 15.

Now, it may be tempting to try and approach this question using trial and error. But it will be more efficient if we can use an algebraic method. We can introduce the letter 𝑥 to represent this positive number. We’re then told that the square of this number, that would be 𝑥 squared, exceeds twice its value, so that’s twice the original number, by 15. We can express this as an equation. The square of 𝑥, 𝑥 squared, is equal to 15 more than twice 𝑥. That’s two 𝑥 plus 15. What we now have is a quadratic equation in 𝑥 because the highest power of 𝑥 that appears is two. We have this 𝑥 squared term here.

Our first step in solving any quadratic equation is to collect all of the terms on the same side. We can do this by subtracting both two 𝑥 and 15 from each side of the equation to give 𝑥 squared minus two 𝑥 minus 15 equals zero. We’ve chosen to group the terms on the left-hand side of the equation because then we have a positive coefficient of 𝑥 squared, which often makes things a little simpler. There are then a number of methods that we may be aware of to solve quadratic equations. They are factoring, using the quadratic formula, or completing the square. If a quadratic equation can be solved by factoring, then this is usually the most straightforward method. So we’ll begin by checking whether it’s possible to factor this quadratic.

The coefficient of 𝑥 squared in this equation is one. So this is one of the more straightforward quadratics to factor. We want to find two linear expressions which multiply to give 𝑥 squared minus two 𝑥 minus 15. The first term in each of these expressions must be 𝑥 because 𝑥 multiplied by 𝑥 gives 𝑥 squared. To complete the parentheses, we’re then looking for two numbers with a specific set of properties. First, their sum must be the coefficient of 𝑥, which is negative two. Second, their product must be the constant term in the equation, which is negative 15.

We can find these two numbers by listing the factor pairs of 15. They are one and 15 and three and five. But we wanted the product to be negative 15, which means we need one of the numbers to be positive and the other to be negative. If we take the second factor pair and we keep the three positive but we make the five negative, then three multiplied by negative five does indeed give a product of negative 15 and three plus negative five gives a sum of negative two. These are the two numbers we need to complete our parentheses then. And so the quadratic factors as 𝑥 plus three multiplied by 𝑥 minus five.

We can, of course, check this by redistributing the parentheses, perhaps using the FOIL method. Multiplying the first terms together, 𝑥 multiplied by 𝑥 gives 𝑥 squared. Multiplying the outer terms, 𝑥 multiplied by negative five is negative five 𝑥. Multiplying the inner terms, three multiplied by 𝑥 is three 𝑥. And finally, multiplying the last terms together, three multiplied by negative five gives negative 15. Collecting the like terms in the center of our expansion gives 𝑥 squared minus two 𝑥 minus 15, which is indeed the quadratic we were trying to factor. So, we have 𝑥 plus three multiplied by 𝑥 minus five is equal to zero.

We then recall that if the product of two numbers or two expressions is equal to zero, at least one of those individual factors must itself equal zero. So we either have 𝑥 plus three equals zero or 𝑥 minus five equals zero. Each of these linear equations can be solved in one step. We solve the first equation by subtracting three from each side to give 𝑥 equals negative three. And to solve the second equation, we add five to each side to give 𝑥 equals five. So we have two solutions to our quadratic equation. Either 𝑥 equals negative three or 𝑥 equals five.

Looking back at the question though, we were using 𝑥 to represent a positive number. So whilst negative three is a valid solution to this quadratic equation, it isn’t a possible value for 𝑥. Our answer then is five, but let’s check this. Five squared is equal to 25. Twice five is 10. And if we add 15, we do indeed get 25. So the square of five does exceed twice its value by 15. Our answer then is five.