Lesson Video: Quadratic Equations: Coefficients and Roots Mathematics • 10th Grade

Video Transcript

In this lesson, we’ll learn how to recognize the relationship between the coefficients of a quadratic equation and its roots. Remember, a quadratic function is of the form 𝑓 of 𝑥 equals 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐, where 𝑎, 𝑏, and 𝑐 are real numbers and 𝑎 is not equal to zero.

Let’s consider a quadratic equation. Now, this is generally of the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. Let’s look at the equation 𝑥 squared minus seven 𝑥 plus 10 equals zero. We can solve this equation, in other words, find its roots, by factoring the left-hand side. And since the coefficient of 𝑥 squared is equal to one, this is fairly straightforward. We look for two numbers whose product is 10 and whose sum is negative seven.

Well, those two numbers are negative two and negative five. So the left-hand side can be factored to 𝑥 minus two times 𝑥 minus five. Then, since the product of the two expressions in parentheses is zero, we can say that either one of those expressions must itself be equal to zero. In other words, 𝑥 minus two is equal to zero or 𝑥 minus five is equal to zero. Solving each equation for 𝑥 gives us two solutions to our equation: 𝑥 equals two and 𝑥 equals five.

But let’s think about the factorization process in reverse. We notice that the roots of the negative of the two numbers that we chose had a product of 10 and a sum of negative seven. So if we define the roots to be 𝑥 sub one and 𝑥 sub two, we can say that negative 𝑥 sub one times negative 𝑥 sub two is equal to 10 and negative 𝑥 sub one plus negative 𝑥 sub two is equal to negative seven. In other words, we’re beginning with a presumption that the coefficients of the quadratic equation actually contain information about its roots. And we can verify that this is true in our example, since our roots are 𝑥 sub one equals two and 𝑥 sub two equals five. Negative two times negative five is 10, and negative two plus negative five is negative seven.

But is this true for any quadratic equation? What is the relationship between the coefficients of any quadratic equation and its roots? Let’s start with a simple quadratic equation, a more general form of the one we just saw. 𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. Say that this factors into some expression 𝑥 minus 𝑥 sub one times 𝑥 minus 𝑥 sub two. Then, the two roots of our quadratic equation are 𝑥 sub one and 𝑥 sub two. But say we were going to distribute the parentheses. We would get 𝑥 squared minus 𝑥 sub two 𝑥 minus 𝑥 sub one 𝑥 plus 𝑥 sub one 𝑥 sub two equals zero. And we could factor the middle two terms to get negative 𝑥 times 𝑥 sub one plus 𝑥 sub two.

Now, let’s compare the coefficients of this equation with that in our original equation. We can say that 𝑐 is equal to 𝑥 sub one 𝑥 sub two and 𝑏 is equal to the negative sum of 𝑥 sub one and 𝑥 sub two. In other words, quadratic equations of the form we’ve given with roots 𝑥 sub one and 𝑥 sub two must satisfy 𝑥 sub one plus 𝑥 sub two equals negative 𝑏 and 𝑥 sub one times 𝑥 sub two equals 𝑐.

Let’s generalize this further, thinking about quadratic equations where their leading coefficient is not equal to one. We can manipulate these to make the coefficient of 𝑥 squared one. So we get 𝑥 squared plus 𝑏 over 𝑎𝑥 plus 𝑐 over 𝑎 equals zero. We’ve just divided through by 𝑎. We can now use our previous results to link the coefficients and roots of any quadratic equation. For a quadratic equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero whose roots are 𝑥 sub one and 𝑥 sub two, they must satisfy the two equations 𝑥 sub one plus 𝑥 sub two equals negative 𝑏 over 𝑎 and 𝑥 sub one times 𝑥 sub two equals 𝑐 over 𝑎. And of course for simple quadratic equations where the coefficient of 𝑥 squared is one, this can be simplified as shown.

What’s really useful is these formula hold for all quadratic equations, even when the roots are complex or are repeated. And the same formulae can be recovered using the quadratic formula. In our first example, we’ll demonstrate how this theorem can help us to find the sum of the roots without actually solving an equation.

Without solving the equation negative three 𝑥 squared minus 16𝑥 plus 63 equals zero, find the sum of its roots.

Remember, if we’re given a quadratic equation of the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero with roots 𝑥 sub one and 𝑥 sub two, these must satisfy the criteria that their sum is equal to negative 𝑏 over 𝑎 and their product is equal to 𝑐 over 𝑎. And so in order to be able to find the sum of the roots of our quadratic equation, we need to identify the values of 𝑏 and 𝑎.

Well, if we look carefully, we can see that the coefficient of 𝑥 squared gives us our value of 𝑎. It’s negative three. The coefficient of 𝑥 gives us our value of 𝑏. That’s negative 16. And 𝑐 is the constant in our equation. It’s 63. We’re looking to find the sum of the roots of this equation. So if we define them, as in our general formula, as 𝑥 sub one and 𝑥 sub two, we know that it’s equal to negative 𝑏 divided by 𝑎. In this case, that’s negative negative 16 divided by negative three. A negative divided by a negative is a positive. So we get 16 over three, and then we make that negative. And so 𝑥 sub one plus 𝑥 sub two is negative 16 over three. And we can now see that, without solving the quadratic equation, we found the sum of its roots. It’s negative 16 over three.

Now, in fact, we’ve demonstrated how to use an equation to find information about its roots. But we can reverse this process. In other words, if we’re given the roots of an equation, we can use the relationship between the coefficient and their roots to find the original equation itself. In this case, it’s generally sensible to start with the most simple form where the coefficient of 𝑥 squared is equal to one, so the equation 𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. Then, if either of the final coefficients 𝑏 and 𝑐 is a fraction, we can simply multiply both sides of the equation by a common denominator to further simplify it. Let’s demonstrate this in our next example.

Given that negative one and negative six are the solutions of the equation 𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, find the values of 𝑏 and 𝑐.

We might recall that, for a quadratic equation of the form 𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, which has roots 𝑥 sub one and 𝑥 sub two, the sum of these roots is equal to negative 𝑏 and their product is equal to 𝑐. Now, we’re told that the roots of the equation are negative one and negative six. So let’s define 𝑥 sub one to be equal to negative one and 𝑥 sub two to be equal to negative six. We know the sum of these two values gives us negative 𝑏. So we have negative one plus negative six equals negative 𝑏. Therefore, negative seven is equal to negative 𝑏, which means 𝑏 is equal to seven.

We can perform a similar process to find the value of 𝑐. This time, it’s the product of the roots. It’s negative one times negative six. Well, negative one times negative six is simply six. So 𝑐 must be equal to six. And whilst not required of us, we could use this information to construct the original quadratic equation. It’s 𝑥 squared plus seven 𝑥 plus six equals zero.

Now, in fact, we can even check our results by solving this equation. If we factor the left-hand side, we get 𝑥 plus one times 𝑥 plus six, which does give us the two solutions and the two roots negative one and negative six, as required. So 𝑏 is equal to seven and 𝑐 is equal to six.

Now, the beauty of what we’ve demonstrated is that it holds for all types of roots. Specifically, it even holds for complex roots. So let’s see what that might look like.

Which quadratic equation has roots 𝑥 is equal to two plus or minus 𝑖? Is it (A) 𝑥 squared minus four 𝑥 plus five equals zero? (B) 𝑥 squared plus four 𝑥 plus five equals zero. Is it (C) 𝑥 squared minus four 𝑥 plus three equals zero? (D) 𝑥 squared plus four 𝑥 plus three equals zero. Or (E) 𝑥 squared minus five 𝑥 plus four equals zero.

Now, we could use some methods such as completing the square or the quadratic formula to solve each of these individual equations. But remember, we have a way to link the coefficients of a quadratic equation with its roots. Specifically, let’s think about a quadratic equation with a coefficient of 𝑥 squared is equal to one. Let’s say that equation has roots 𝑥 sub one and 𝑥 sub two and the coefficient of 𝑥 is 𝑏 and the constant term is 𝑐. The sum of the roots is equal to negative 𝑏, and the product of the roots is equal to 𝑐.

So, given information about our two roots, we can find the value of negative 𝑏 by adding them and we can find the value of 𝑐 by multiplying them. Specifically, negative 𝑏 is two plus 𝑖 plus two minus 𝑖. And of course we can add pairs of complex numbers by just adding the real parts and then separately adding the imaginary parts. So that’s two plus two plus 𝑖 minus 𝑖. But 𝑖 minus 𝑖 is simply zero, so that gives us a real value of four. And since negative 𝑏 is equal to four, we can say 𝑏, which is going to be the coefficient of 𝑥 in our equation, must be equal to negative four. And this is great, in fact, because we can instantly disregard option (B) and (D) from our solutions. They have a coefficient of 𝑥 of positive four.

Next, we’ll find the product of the roots to find the value of 𝑐. That’s two plus 𝑖 times two minus 𝑖. Distributing the parentheses, and we get four minus two 𝑖 plus two 𝑖 minus 𝑖 squared. Negative two 𝑖 plus two 𝑖 is zero. And since 𝑖 squared is negative one, this simplifies further to four minus negative one, which is simply five. So we have a quadratic equation where the coefficient of 𝑥 squared, we said, is one; the coefficient of 𝑥, 𝑏, is negative four; and the constant term 𝑐 is five. So our equation is 𝑥 squared minus four 𝑥 plus five equals zero. And that, we see, corresponds to answer (A).

So now that we’ve looked at several examples that demonstrate the link between the coefficients and roots of a quadratic equation, let’s look at how we can solve more complicated problems involving unknowns.

If 𝐿 and 𝑀 are the roots of the equation 𝑥 squared plus 10𝑥 plus nine equals zero, what is the value of 𝐿 squared plus 𝑀 squared?

Now, one way we can answer this problem would be to solve the original equation. But we are in fact given information about the roots, and we’re given the equation itself. And so we recall that, given a quadratic equation with a coefficient of 𝑥 squared is one, and if it has roots 𝑥 sub one and 𝑥 sub two, the sum of these roots is negative 𝑏 and the product of these roots is 𝑐. So our roots are 𝐿 plus 𝑀. So what’s their sum?

Well, the coefficient of 𝑥 here, the value of 𝑏, is 10. So 𝐿 plus 𝑀 must be equal to negative 10. Similarly, the product of these roots must be equal to the constant term, which is nine. So we have two equations: 𝐿 plus 𝑀 equals negative 10 and 𝐿𝑀 equals nine.

And we might not know where to go next, but we are trying to find the value of 𝐿 squared plus 𝑀 squared. So let’s imagine what might happen if we square both sides of this first equation. When we do, we get 𝐿 plus 𝑀 all squared equals negative 10 squared. But then if we distribute the left-hand side, we get the expression 𝐿 squared plus two 𝐿𝑀 plus 𝑀 squared. And we can see that that corresponds in some respects to the value that we’re looking for. And in fact negative 10 squared is 100. So this is equal to 100.

Since we have two 𝐿𝑀 and we know 𝐿𝑀 is equal to nine, we can rewrite two 𝐿𝑀 as two times nine, which is 18. And so we have an equation 𝐿 squared plus 18 plus 𝑀 squared equals 100. Then, we solve for 𝐿 squared plus 𝑀 squared by subtracting 18 from both sides. 100 minus 18 of course is 82. And so the value of 𝐿 squared plus 𝑀 squared is 82.

We’ve now demonstrated in some detail how coefficients of a quadratic equation contain information about the sum and product of its roots. Let’s recap the key concepts.

In this lesson, we learned that, for a quadratic equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, if it has roots 𝑥 sub one and 𝑥 sub two, then the sum of these values are negative 𝑏 over 𝑎. And similarly, the product of the roots is 𝑐 over 𝑎. Of course, if 𝑎 is equal to one, if the coefficient of 𝑥 squared equals one, then this simplifies to 𝑥 sub one plus 𝑥 sub two equals negative 𝑏 and 𝑥 sub one times 𝑥 sub two equals 𝑐. And of course we can reverse this idea. We can use information about the roots to find coefficients of a quadratic equation and, hence, find the equation itself.