Hello wonderful mathematics people, this is Anna Cox from Kellogg Community College. Quadratic equation AX squared plus BX plus C = 0. The quadratic formula is X equal negative B plus or minus the square root b ^2 -, 4 AC all over 2A. The discriminant is the inside of the square root, hence the b ^2 -, 4 AC portion. The discriminant tells us what types of solutions to expect, whether they be real numbers, imaginary numbers, rationals, or irrationals. So if the discriminant equals 0, we'd really end up with something that says negative b ± sqrt 0 / 2 A. So we'd have negative b / 2 A occurring twice. So it actually is considered a double root, and it's a rational number, so it's one depth double rational solution. If the discriminant is less than 0, IE the inside would be a negative number. The square root of a negative is imaginary, so we'd have two imaginary solutions. If the discriminant is greater than 0, then we have two real solutions, and if the b ^2 - 4 AC is a perfect square, then we have two rational solutions, otherwise we have two irrational solutions. So let's look at some examples. If we have X ^2 - 7 X plus 5 = 0, we look at the b ^2 - 4 AC portion and we get -7 ^2 - 4 * 1 * 5 49 - 20 or 29 so 29 tells me we're going to have two irrational solutions. If we look at this next one, b ^2 - 4 * A * C so 16 - 24 or -8 So this tells me we're going to have two imaginary solutions. Four X ^2 - 8 X -5. So our b ^2 - 4 * a * C so -8 ^2 is 64 + 80 is going to be 144. And that's actually a perfect square that's 12 ^2. So this is going to be two rational solutions. Our next one negative 24 ^2 - 4 * 16 * 9. When we compute that, we're going to get 0. So this tells US1 double rational root. The next type of problem is if we're given the actual solution, so X equal -5 and X equal 4, we want to come up with the quadratic. So we're going to add the number to one side so that we have 0 on each side of those two individual equations. Now what we're going to do is we're going to actually multiply the left sides together, multiply the right sides together. If we foil out the left side, we get X ^2 -, 4 X plus five X -, 20 equaling 0. So our answer is going to be X ^2 + X - 20 = 0. If we found the solutions, they would be -5 and four. So X equal 2/3, X equal -8 three X equal 2, three X - 2 equals zero, X + 8 = 0. We're going to multiply the three X - 2 and the X + 8 and on the other side 0 * 0 zero. So when we foil it out, three X ^2 + 24 X -2 X -16 or three X ^2 + 22 X -16 = 0. If we have X equaling root 5 and X equaling negative root 5. Same concept, we're going to take it all to one side. So X minus root 5 = 0, X plus root 5 = 0. So when we multiply X minus root 5 and X plus root 5, we're going to get X ^2 plus root 5X and minus root 5X. So that plus root 5 and minus root 5 are going to cancel. So when we finish foiling it and combine like terms, we get X ^2 - 5 = 0. If we have X = 7 I and X equal negative seven, 9 - 7 I equals zero, X + 7 I equals 0. When I multiply these, we're going to get X ^2 + 7 nine minus seven 9 - 49 I squared. So X ^2 the seven nine and the negative 7/9 cancel I ^2 is really -1 negative 49 * -1 is plus. So that's our solution. The next one, we're going to get X = 3 plus square to two, X = 3 minus square to two. Take everything to one side. So X -, 3 minus root two, X -, 3 plus root 2. When we foil this, actually we're not going to foil. We have to distribute here because we have 3 terms times 3 terms. If we multiply it, we'd get X ^2 -, 3 X plus root 2X. We'd get negative three X + 9 -, 3 root 2. We'd get a negative root 4A, positive 3, root 2, and a negative root 2. XI just did that out of order. Let's see, X * X is X ^2 -3 X, positive root 2, X -3 X +9, negative 3, root two -2 root X +3, root 2, and a negative root 4. So when we combine this, we get X ^2 - 6 X. The roots are going to cancel 9 -, 2 or seven, and that's all going to equal 0. So that would be our final answer for X ^2 -, 6 X +7. If we factored that, we'd get 3 plus square to 2 and 3 -, sqrt 2. Thank you and have a wonderful day.