Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Quadratic equation AX squared plus BX plus C = 0. We're going to develop a formula that can solve anything of this form. We're going to start by dividing everything through all the coefficients through by the leading coefficient. So we're going to have a / A * X ^2 + B / a * X plus C / a = 0 / a. So a / a is really just one the b / a times the X 0 / a is 0, but we're going to take that C / A to the other side. So we're going to get negative C / a. This next step, we're going to take half of the coefficient on the linear term and square it. So we're going to have b / 2 A and we're going to square it. This is how we complete a square. And if we do it to one side, we need to be able to do it to the other. So b / 2 a quantity squared plus that negative C / a. So the left side turns into X ^2 + b over AX plus b ^2 / 4 A squared. The right side, the b ^2 / 4 A squared plus or minus is really just the same thing as subtraction of C / a. The left side is now a perfect square, so we're going to be able to factor it. That's going to factor into X + b / 2 A. If you think about that b / 2 A times b / 2 A is b ^2 / 4 A squared and b / 2 A from the outer term plus another b / 2 A from the inner term. When we foil would give us 2B over 2A's so that two and the two in the denominator would cancel, leaving us the b / a coefficient. So now on the right side, I want to get a common denominator. The common denominator is going to be 4A squared. So the C / a has to get multiplied by 4A to get a 4A squared. And if we do the bottom, we have to do the top. So the top's going to turn into four AC all over 4A squared. So X + b over 2A squared is going to equal b ^2 -, 4 AC all over 4A squared. Now how do we get rid of a square? We're going to square root it. And if I square root one side, we need to square root the other. So I'm going to get X + b / 2 A equaling positive and negative square root b ^2 - 4 AC over sqrt 4 A squared. I want X by itself, so I'm going to subtract that negative b / 2 A+ or minus the square root b ^2 - 4 AC. Now sqrt 4 is just two and sqrt a ^2 is a in the denominator. So now we have the common denominator of 2A. So X equal negative B plus or minus the square root b ^2 -, 4 AC all over 2A. This is called the quadratic formula and that's what we were trying to develop. So given any quadratic equation, IE the quadratic equation was the ax squared plus BX plus C = 0, we can now go straight to the quadratic formula and not have to factor. Now there are reasons why factoring is easier and there's reasons why quadratic formula is easier. The quadratic formula will allow us to solve things that are not factorable. So let's look at a few examples. If we have two X ^2 -, 3 X +5, we get our A equaling 2, our B equaling -3, and our C equaling 5. We're going to stick it into the formula we just developed. AX equal negative B plus or minus the square root b ^2 -, 4 AC all over 2A, so X is going to equal the opposite of B. The opposite of a -3 is a positive 3 plus or minus the square root b ^2 -3 ^2 - 4 * a * C all over 2 * a, so +3 plus or minus the square root 9 - 40 all over 4. We need to simplify that 9 -, 49 - 40 is going to give us a negative 31 and we don't leave a negative inside the square root. So we're going to have 3 ± I sqrt 31 / 4 or if we put that in a plus BI form, it would be 3/4 ± sqrt 31 fourths times I. Now in the next example, if we don't have it equal to 0, we're going to just subtract that 6X so that we get everything on one side equaling TO0A equal 1B equal -6 C equal 3. So X equals the opposite of B plus or minus the square root b ^2 - 4 * a * C all over 2 * a, so the opposite of a -6 is +6. Negative 6 ^2 is 36 - 4 * 1 * 3 or 12 / 2, so 36 - 12 is 24. Now 24. We can simplify 24 if we think about it is 4 * 6 or 2 * 2 * 2 * 3. So we'd have 6 ± sqrt 2 ^3 * 3, and that index understood of two goes into the exponent of three one time with one leftover. So the leftover stays underneath the radical. The cold quantity comes outside. So we'd get 6 ± 2 square roots of 6 / 2. Now the numerator has a 2IN common, so we're going to factor out of two and we get 3 ± sqrt 6 left over 2. Finally, we have a monomial over a monomial, so the two in the numerator and the two in denominator will cancel. And our final answer is going to be 3 ± sqrt 6. So X equal 3 ± sqrt 6. Let's look at one that has to do some distribution before we can even use quadratic formula. So here we're going to distribute. We get seven X ^2 + 14 X +5 equaling three X ^2 + 3 X. We're going to take all of the terms to one side, so we can get one side equaling 0. So seven X ^2 - 3 X squared is four X ^2 14 X -3 X is eleven X + 5 = 0. So here our A is going to be 4, our B is 11, and our C is five. X equals the opposite of B plus or minus the square root b ^2 -, 4 * a * C all over to A so -11 plus or minus. The square of 11 is 121 -, 80 all over eight 121 -. 80 is going to give U.S. 41 and 41. We actually can't simplify, so -11 ± sqrt 41 / 8 is our final answer there. What if we have a cube? Well, the exponent's going to tell us how many solutions to expect. So if it's a cube, we should expect 3 solutions. On the other problems, they were squares, so we expected 2 solutions. If we jump back for just a second, we have a -11 + sqrt 41 / 8 as one answer and -11 -, sqrt 41 / 8 as your second solution. So there were two solutions on that one as there were on both the previous. That plus and minus means there's one answer that's a plus and one answer that's a minus. Well, this one starts out with being a formula that hopefully we know, and it's the sum of cubes. Remember, if we have a ^3 + b ^3, that's A + B * a ^2 minus AB plus b ^2. So this formula over here is going to factor into X + 2 times X ^2 - 2, X plus 4 = 0. Now from this first parenthesis, we know that X + 2 is going to equal 0 or X = -2. The second parenthesis is now a quadratic. When we set it equal to 0, it doesn't factor. If it had, we would have factored it in our formula originally. So now we're going to use the quadratic formula. Here we know that our A equal 1, our B equal -2, and our C equal 4. So X equals the opposite of B plus or minus the square root b ^2 -, 4 * a * C all over to A. So X equal +2 plus or minus. The square root -2 ^2 is 4 - 4 * 1 * 4 negative 16 / 2. So that's going to equal 4 - 16 as -12. Now -12 can simplify. Remember the negative is going to come out as an I. So if we prime factor 12 two and six, 6 is 2 * 3, so we'd get 2 ± I sqrt 2 ^2 * 3, and the square root of that 2 ^2 is just really 2, so we get 2 I root 3 / 2. Now that numerator has a 2IN common, so we can factor out A2 to make it into a monomial. If I take out a 2, I get 1 ± I square roots of three all over 2 monomial over monomial. So those twos can cancel and our final answer is going to be X equal 1 ± I square roots of three. So this equation really has three final answers -2 1 + I square roots of three, and 1 -, I square roots of three. The last type of example I'm going to do is if it's in rational expression form. So we're going to start by finding our common denominator of X * X + 4, and we're going to always state that X can't equal whatever would make the denominator go to 0. So X isn't zero and X isn't -4. If we multiply through by the common denominator X * X + 4 times that 7 / X plus that X * X + 4 times the 7 / X + 4 and the X * X + 4 times the one. So in this first one, the XS are going to cancel and we're going to get left 7 * X + 4 and the next 1. The X plus fours are going to get cancel and we're going to get left 7X and on the right side when we multiply we're going to have X * X + 4. So distributing this out seven X + 28 + 7 X equals X ^2 + 4 XI. Personally always find it easier if my squared term is positive. So I'm going to move the 14X and the 28 over to the right hand side. So I'm going to get 0 equal X ^2 - 10 X -28 A equal 1B equal -10 C equal -28. So X equals the opposite of B plus or minus the square root b ^2 - 4 * a * C all over 2A. So +10 ± sqrt 100 + 112 all over 2. So 10 ± sqrt 212 / 2. Now 212 if we thought about prime factoring it, a four will go in. Four goes into 21 five times and four goes into twelve. 3 * 53 is prime, so we're going to get 2 * 2 * 53. So sqrt 212 is going to really simplify to 10 plus or minus sqrt 2 ^2 * 53, and that 2 ^2 is going to be just plain old 2. When I take the square root of it, we're going to realize that the numerator can have a two that factors out. So we get 5 ± sqrt 53 times the 2 / 2. Now they're both monomials in the numerator and denominator, so they'll cancel, leaving me 5 ± sqrt 53. That doesn't conflict with what our X can't equal. So we have two answers. X is 5 + sqrt 53 and 5 -, sqrt 53. Thank you and have a wonderful day.