# Question Video: Using the Sign of the Discriminant to Determine the Number of Complex Roots of a Quadratic Equation Mathematics • 9th Grade

How many nonreal roots will a quadratic equation have if its discriminant is negative?

03:09

### Video Transcript

How many nonreal roots will a quadratic equation have if its discriminant is negative?

Before we can answer this question, let’s begin by defining some of the key terms. A quadratic equation is of the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. In this example, we’re going to be thinking purely about values for 𝑎, 𝑏, and 𝑐, which are real numbers. But for any quadratic equation, 𝑎 cannot be equal to zero. The roots of this quadratic equation are the solutions or the values of 𝑥 that satisfy 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. And then, the question asks us how many nonreal roots will it have if the discriminant is negative.

So, what do we mean by the discriminant? Well, to answer this, we recall the quadratic formula. This tells us that the solutions to the equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero are 𝑥 equals negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎. The discriminant is this expression inside the square root. We use Δ to represent it, and it’s 𝑏 squared minus four 𝑎𝑐.

Now, it follows that if the discriminant, if 𝑏 squared minus four 𝑎𝑐, is positive, when we take the square root, we get a real solution. And in particular, if the discriminant is positive, the square root is a nonzero real number. This means that when we have a positive discriminant, there are two real solutions to the quadratic equation. In other words, there are two real roots. If, however, the discriminant is equal to zero, we then take the square root of zero, which is zero. This means that our solutions to 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero are simply negative 𝑏 over two 𝑎. So, that’s actually one solution. So, when the discriminant is zero, we get one real root.

But what about if the discriminant is negative, as in this question? Well, if it’s negative, we’re taking the square root of a negative number, which gives us an imaginary solution. This means there are, in fact, no real roots. We instead get complex roots. But how many complex roots do we get? Well, let’s suppose that when we take the square root of 𝑏 squared minus four 𝑎𝑐, we get some value 𝑑 times 𝑖, where 𝑑 is a constant, a real constant, and 𝑖 is the imaginary number, which is the solution to the equation 𝑖 squared equals negative one.

In this case, the solutions to our quadratic equation are 𝑥 equals negative 𝑏 plus or minus 𝑑𝑖 over two 𝑎. We can separate this into the two distinct solutions 𝑥 sub one and 𝑥 sub two. These are not real, but they are complex solutions to the quadratic equation. They are negative 𝑏 plus 𝑑𝑖 over two 𝑎 and minus 𝑏 minus 𝑑𝑖 over two 𝑎, respectively. We see that there are two complex solutions to the quadratic equation, even though the discriminant is negative.

And so, the answer is two. A quadratic equation will have two nonreal roots if its discriminant is negative.