Hey everyone. So we've just finished learning about all of the different methods of solving quadratic equations but we're gonna take a quick detour here to talk about the discriminant. Now the discriminant might feel a little bit random here and you might be wondering why you need to learn about it, but the reason I'm telling you about it at all is because you're going to be asked a very specific type of question about how many real or imaginary solutions a quadratic equation has and you're not gonna wanna have to go through fully solving it just to answer that it has one real solution. So that's where the discriminant comes in. So I'm gonna show you exactly how to use the discriminant to answer this type of question really quickly. Let's go ahead and dive in.
The discriminant is just the expression underneath the radical in the quadratic formula. So it's just b2-4∗a∗c. And the sign of the discriminant is going to tell us the number and the type of solutions. So if the discriminant is positive, that means that my quadratic equation is going to have 2 real solutions. If it's 0, that tells me that I only have one real solution. And if it's negative, that tells me that I actually have no real solutions and I have 2 imaginary ones. Whether it's positive, 0, or negative, I can see easily and quickly whether it has 2 real solutions, one real solution, or none at all.
So let's go ahead and look at a couple of different quadratic equations and identify the number and the type of solutions they have. So looking at my first one here, I have 2x2+3x-2=0. Now if I want to calculate my discriminant here, let's go ahead and label a, b, and c to make it a bit easier to plug in. So a here is 2, b is 3, c is negative 2. Plugging that all into my discriminant formula, I have 32-4∗2∗(-2). Now simplifying that, 32 is 9 minus 4∗2∗2 is 8 times -2, and this becomes 9 + 16, which gives me 25. Now this is a positive number, so that tells me that I have 2 real solutions. So this quadratic equation has 2 real solutions.
Let's look at another example. So I have 4x2+x+2=0. Let's go ahead and label a, b, and c. Here, a is 4, b is this 1, I have an invisible one there, and c is 2. Now calculating the discriminant, I have 12-4∗4∗2, just plugging everything in. So simplifying this, 12 is just 1 minus 4∗4∗2 is 16 times 2, and this becomes 1 minus 32 which gives me -31. So I am left with a negative number, which tells me that I have no real solutions and I have 2 imaginary ones. So here I have 0 real and 2 imaginary solutions.
Let's take a look at one final example here. So I have x2-10x+25=0. Now going ahead and labeling a, b, and c, a here is 1, b is -10. Make sure to look at that sign, and c is 25. So plugging this into my discriminate formula, I have -10)2-4∗a∗c. Now simplifying this, -102 is 100 minus 4∗1∗25 is just 4, so this is really just 4 times 25, which is 100. And 100 minus 100 is going to give me 0. So finally here, I have that my discriminate is 0, which is the last option that I have for my, discriminate. And that tells me that I only have one real solution, And that's all there is to the discriminate. Now whenever you're asked this question, you can do it really quickly. Let me know if you have any questions.
