Lesson Video: Solving Quadratic Equations: Quadratic Formula Mathematics

Video Transcript

In this video, we will learn how to solve quadratic equations using the quadratic formula. This is a standard formula that we can apply to help us find the solutions to certain types of quadratic equations. Suppose we have the general quadratic equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, where 𝑎, 𝑏, and 𝑐 are constants, that is, just numbers, and 𝑎 isn’t equal to zero. The reason for this is that if 𝑎 was zero, we would just have a linear equation. The quadratic formula tells us that the solutions to this quadratic equation, if they exist, are given by 𝑥 equals negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎.

Now, when I said if they exist, I mean that there are certain quadratic equations which actually have no real-valued solutions. And trying to use this formula won’t work. Can you see from the formula how this might be the case? Well, we can see that the formula includes the square root of 𝑏 squared minus four 𝑎𝑐. And we also know that whenever we take a square root, we need the value we’re attempting to square-root to be either positive or zero. So, if the expression underneath the square root — that’s 𝑏 squared minus four 𝑎𝑐 — turns out to have a negative value, we’d be trying to find the square root of a negative value, which won’t work. Now, you don’t need to worry about this now because the quadratic equations you’re asked to solve at this level won’t have this problem. But if you go on to study maths at a higher level, you’ll look at this in a little bit more detail.

Now, if you’re interested in where this formula comes from, it can actually be derived by rearranging the equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero to make 𝑥 the subject. And I’ve written the first couple of steps in this process here. One of the first key steps is that we need to factor by 𝑎. That’s the coefficient of 𝑥 squared. And, as we said, 𝑎 must be nonzero. We can then divide both sides of the equation by this factor. The next key step is that we need to use the technique of completing the square. And I’ve written the first stage of this here. If you fancy a challenge, it’s a great exercise to practice your algebra skills to work through the remainder of this method.

Now, it’s also important to point out that if a quadratic equation can be solved by factorizing, then using the quadratic formula will still work and will give exactly the same result. Although it will probably be more effort for you and there’s more chance of making a mistake. Where the quadratic formula is really useful is when a quadratic equation cannot be factorized. The focus of this video will be gaining confidence in applying this formula. Now, depending where you are in the world and the requirements of your local exam board, you probably need to know this formula off by heart. Don’t worry if it looks a little bit scary at this stage. The more times you use it, the more natural it will become. And you’ll be surprised how quickly you learn the formula. Don’t believe me? Well, let’s try some examples.

Find the solution set of the equation five 𝑥 squared minus seven 𝑥 minus 32 equals zero, giving values to three decimal places.

So, we have the quadratic equation and we’re asked to find its solution set, which is just another way of saying solve the equation. We’re told that we need to give values to three decimal places. So, this is a big clue that the quadratic equation we’ve been given won’t be able to be solved by factoring. And so, we need to use another method. This is where we need to use the quadratic formula, which we should make sure we learn off by heart. The quadratic formula tells us that if we have the general quadratic equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. Then the solution set or roots of this equation, if they exist, are given by 𝑥 equals negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎.

In order to apply this formula then, what we need to do are determine the values of 𝑎, 𝑏, and 𝑐. 𝑎 is the coefficient of 𝑥 squared. So, for our equation, that’s five. 𝑏 is the coefficient of 𝑥. So, for our equation, that’s negative seven. And we must make sure that we include the negative sign. 𝑐 is the constant term. In our equation, that’s negative 32. Now, all we need to do is substitute these values of 𝑎, 𝑏, and 𝑐 into the quadratic formula. So, we have 𝑥 equals negative 𝑏 — that’s negative negative seven — plus or minus the square root of 𝑏 squared, so negative seven squared, minus four 𝑎𝑐 — that’s minus four times five times negative 32. And this is all over two 𝑎. That’s two times five. You can break the formula down into stages if you wish, or you can substitute straight into the formula. But you must be careful.

Let’s now look at simplifying. We have negative negative seven. So that’s seven. And in the denominator, two times five is 10. Let’s now look at the square root. Firstly, we have negative seven squared, and negative seven squared is equal to 49. If you’re using a calculator to help with this, it’s really common to make a mistake when finding this value. Because what a lot of people do by mistake is they type negative seven squared into their calculator, exactly as I’ve written in orange here.

Now, when you do this, your calculator will give negative 49, but that doesn’t mean your calculator has made a mistake. It means you’ve made a mistake in the question you’ve asked it. Negative seven squared means negative one multiplied by seven squared. And if you recall BIDMAS or the order of operations, we work our indices or powers before multiplication. So, negative seven squared means negative one multiplied by seven squared, which is negative one multiplied by 49, giving negative 49. What we should’ve typed into our calculator was negative seven enclosed in brackets squared. So that ensures that our calculator is taking a negative value first and then squaring it. This is a really common mistake, and one you need to make sure you watch out for.

Continuing under the square root then, we are subtracting four times five times negative 32. Now, four times five times negative 32 is negative 640. So we are subtracting negative 640, which means overall we’re adding 640. But again, you need to be really careful with the negative signs here. Simplifying under the square root, 49 plus 640 is 689. So, we have that 𝑥 is equal to seven plus or minus the square root of 689 over 10.

Now, we’re asked to give our values as decimals, so we now need to use our calculator to evaluate each of these. The first root or value in our solution set is seven plus the square root of 689 all over 10, which is equal to 3.32488 continuing. The second root or second value in our solution set is seven minus the square root of 689 all over 10. And that’s equal to negative 1.92488. We’re asked to give the values to three decimal places, and in each case, the fourth decimal place is an eight. So, we round up. We can then give our two roots in set notation. The solution set to this equation is 3.325, negative 1.925.

The important stages in this question then were correctly identifying the values of 𝑎, 𝑏, and 𝑐 from our quadratic equation, including their signs, and then carefully working with the signs when applying the quadratic formula. In our next example, we’ll see how to apply the quadratic formula to an equation which may not look like a quadratic equation at first glance.

Find the solution set of the equation two 𝑥 minus five equals six over 𝑥, giving values to three decimal places.

Now, at first glance, we may not know how to solve this equation as it involves a reciprocal term, six over 𝑥. There’s a trick, though, that we need to spot, which is that if we multiply the entire equation by 𝑥, this will eliminate the fraction. On the left-hand side, we’ll have the 𝑥 multiplied by two 𝑥 minus five. And on the right-hand side, 𝑥 multiplied by six over 𝑥. We can distribute the parentheses or expand the brackets on the left-hand side to give two 𝑥 squared minus five 𝑥. And on the right-hand side, 𝑥 multiplied by six over 𝑥 simplifies to six. And we’ve eliminated the fraction. We now see that what we have is a quadratic equation. And so, our next step is to collect all the terms on the same side.

We can do this by subtracting six from each side of the equation, giving two 𝑥 squared minus five 𝑥 minus six is equal to zero, a quadratic equation in its most easily recognizable form. Now, we could perform a quick check to see whether this quadratic equation can be solved by factoring. But as we’re told in the question to give the solution set to three decimal places, that’s a huge clue that it can’t. So, we’re going to need to apply the quadratic formula instead. This tells us that, for the general quadratic equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. Then the solution set or the roots of this quadratic equation are given by 𝑥 equals negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎.

Now, it’s important to notice that the quadratic formula uses a quadratic equation in the form where all the terms are on one side, and we have zero on the other. So, when we’re identifying the values of 𝑎, 𝑏, and 𝑐 to substitute into the formula, we must use our quadratic equation in its rearranged form, not in its earlier form of two 𝑥 squared minus five 𝑥 equals six. If we were to use the earlier form, then we would have the sign for the value of 𝑐 incorrect, which would lead to an incorrect solution. Let’s determine then the values of 𝑎, 𝑏, and 𝑐 for our quadratic equation. 𝑎 is the coefficient of 𝑥 squared. That’s equal to two. 𝑏 is the coefficient of 𝑥, which is equal to negative five. And finally, 𝑐 is the constant term, which is equal to negative six. Remember, we must include the sign with each of these values.

Next, we need to substitute the values of 𝑎, 𝑏, and 𝑐 carefully into the quadratic formula. We have 𝑥 equals negative 𝑏 — that’s negative negative five — plus or minus the square root of 𝑏 squared, negative five squared, minus four 𝑎𝑐 — that’s minus four times two times negative six — all over two 𝑎, which is two times two. Now, we simplify, being very careful with the negative signs. Negative negative five is simply five. And in the denominator, two times two is four. Within the square root, negative five squared is 25.

Be very careful here, particularly if you’re using a calculator. A common mistake is to write that as negative 25. We’re then subtracting four times two times negative six. So, we’re subtracting negative 48. Within the square root, 25 minus negative 48 is the same as 25 plus 48, which is 73. So, we have that 𝑥 is equal to five plus or minus the square root of 73 over four. And that’s our exact answer in terms of surds.

The question though asked us to give our values to three decimal places. So, we now need to evaluate these on a calculator. Our first value is 𝑥 equals five plus the square root of 73 all over four, which is 3.38600. Our second value is five minus the square root of 73 all over four, which is negative 0.88600. Rounding these values to three decimal places then gives the solution set of 3.386, negative 0.886.

There are two important points to remember from this question. Firstly, we must make sure our quadratic equation is in the correct form with all the terms on the same side of the equation before identifying the coefficients 𝑎, 𝑏, and 𝑐. And secondly, we must be really careful with negative values in our quadratic formula. Let’s consider one final example, which will reveal something interesting about the relationship between the sum of the roots of a quadratic equation and its coefficients.

The sum of the roots of the equation four 𝑥 squared plus 𝑘𝑥 minus four equals zero is negative one. Find the value of 𝑘 and the solution set of the equation.

So, we’re looking, first of all, to find the value of 𝑘, which is a missing coefficient in this quadratic equation. It’s the coefficient of 𝑥. We’re going to need to use the quadratic formula to find expressions for the roots of this equation, which will be in terms of 𝑘. And we recall that, for the general quadratic equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, its roots are given by 𝑥 equals negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎.

Let’s determine the values of 𝑎, 𝑏, and 𝑐 for this quadratic equation then, which is in the correct form with all the terms on the same side. The value of 𝑎 is the coefficient of 𝑥 squared. So 𝑎 is equal to four. The value of 𝑏 is the coefficient of 𝑥. So 𝑏 is equal to this unknown value 𝑘. And the value of 𝑐 is the constant term, so 𝑐 is equal to negative four. Substituting into the quadratic formula then, we have that 𝑥 is equal to negative 𝑘 plus or minus the square root of 𝑘 squared minus four multiplied by four multiplied by negative four all over two multiplied by four. In the denominator, two times four is eight. And within the square root, four times four times negative four is negative 64. And subtracting negative 64 is the same as adding 64. So, we have negative 𝑘 plus or minus the square root of 𝑘 squared plus 64 all over eight.

Now, we’re told this key piece of information, that the sum of the roots of this equation is negative one. Adding the expressions for our two roots then, and we have the equation on the screen. Now, as both of these fractions have a common denominator of eight, we can actually combine to a single fraction. And what we notice is that there’s actually quite a lot of simplification. We have plus the square root of 𝑘 squared plus 64, and then minus the square root of 𝑘 squared plus 64. So, these two terms will cancel each other out. We’re left with the far simpler equation negative two 𝑘 over eight is equal to negative one, which we can solve by multiplying by eight and dividing by negative two to give 𝑘 equals four.

In fact, this is illustrative of a general and really useful result. When we added our two roots together, the part under the square root canceled out due to them having different signs. So, what we were left with, the contribution from each root to the sum, was just negative 𝑏 over two 𝑎. We therefore summed two lots of this, giving negative two 𝑏 over two 𝑎, which simplified to negative 𝑏 over 𝑎. In our case, we found that 𝑘 equals four. So, the value of 𝑏 in our quadratic equation is four. And so, negative 𝑏 over 𝑎 gives negative four over four, which is negative one, the correct sum for the roots. What this has shown though is that, in general, the sum of the roots of a quadratic equation is equal to negative 𝑏 over 𝑎. And we can quote this as a general result.

Now that we know the value of 𝑘, we need to determine the values of 𝑥. We have negative four plus or minus the square root of four squared plus 64 over eight. We can simplify the surd here. The square root of 80 is equal to four root five. And then dividing through by a common factor of four gives our simplified solutions as negative one plus or minus the square root of five over two. So, we’ve completed the problem. The value of 𝑘 is four, and the solution set of the equation in surd form is negative one minus root five over two, negative one plus root five over two.

Let’s summarize then what we’ve seen in this video. The quadratic formula can be used to solve quadratic equations of the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 is equal to zero. The solution set or roots of this equation, if they exist, are given by 𝑥 equals negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎. In our final example, we also saw that the sum of the roots of the quadratic equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero is negative 𝑏 over 𝑎, which we can use as a general result.