[Music] video we're going to look at how to solve quadratics by using the quadratic formula now whenever you've got a quadratic uh there's different ways to solve it as we've seen how drawn it looking at the graph by factorizing it and solving it that way using triel Improvement now another way of solving It Is by using the quadratic formula the quadratic formula is this now it looks really complicated but don't worry it's not too bad okay uh for the gcsc test it's normally given to you anyway and it's at a level then you would have to learn it okay so the quadratic formula is X = B+ orus the S < TK of B ^2 - 4 a c all ID by 2 a okay um there's a there's some cool videos in YouTube which help you remember this okay there's a song quadratic formula song um if you're in a class I've probably played it already um but this is the quadratic formula now obviously you need to know what the a the B and C stand for so if you've got a quadratic something like this a x^2 + b x + C and equal 0 the a is the coefficient of the x² term the B is the coefficient of x and C is the constant at the end okay so the a is the number in front of the X squ the B is the number in front of the X and the C is the number at the end okay and then you substitute these values for a b and c into this formula and you'll get your solutions for X okay so let's do question question using a quadratic formula so we're going to solve x^2 + 8 x + 15 = 0 using the quadratic formula so here's the quadratic formula x = b plus orus the square < TK of B ^2 - 4 a c all ID 2 a so uh first of all let's label our a b and c so here a is the coefficient of X2 so it's one okay remember there's really a one here in front of the X squ so a equal 1 B is the coefficient of x which is 8 and C is the constant which would be + 15 or 15 okay so let's then substitute those into the quadratic formula so the quadratic formula is B so - 8 so x = - 8 plus or minus the square root of b^ 2 so that's 8^ s minus now I always put this 4 A Part in Brackets it comes in useful whenever you've got negatives and so on for a b and c you'll see why later so I always put this in Brackets so four * a which is 1 * C which is 15 and then we're going to divide that by 2 a so 2 * 1 so let's work out what that is so -8 plus or minus the square Ro T of 82 is 64 minus 4 * 1 is 4 * 15 is 60 so - 60 all ID 2 when you work out this part here uh 64 take 60 is 4 so you're going to get - 8 plus or minus Square < TK of 4 all / 2 now let's work out this part so you've got - 8 plus orus sare < TK of 4 well square of 4 is 2 so then you get equals - x - 8 plus or - 2 / 2 now you notice you got this plus or minus that means you've got two solutions uh remember if you wanted to solve this quadratic here you could have factorized it and you would have got two solutions actually this one does factorize have done that on purpose so what I do at this point is whenever you sort of simplify this down as fully as possible like we have separate it okay so you've got x = - 8 + 2 / 2 or x = - 8 - 2 / 2 so - 8 + 2 is - 6 / 2 would then be Min -3 so one solution is -3 or going to this one - 8 to2 would be - 10 half of that is - 5 so our Solutions are x = -3 or X = -5 and that's it so that's the quadratic formula and that's how we can use it to solve uh quadratic let's have a look at typical exam question now solve the equation x^2 - 10 x - 5 = 0 given your answer to two decimal places giving your answers to two decimal places now there's a clue in the question here that we're going to be using the quadratic formula and it's the fact we have to give our answer to two decimal places normally whenever you factorize it normally you you're getting an answer that is you know an integer or to one decimal place so uh first thing is let's label our a b and c so a equals the number in front of the x or coefficient of x sorry the number the number in front of the x squ or the coefficient of x squ which would be one the B is going to be equal to -10 and our c is equal to - 5 so um what we're now going to do is then substitute these into the formula so we're going to get x equals now it says b now B is already minus uh - 10 now so Min - - 10 would then be 10 plus or minus the square Ro T of b s so it's going to be - 10^ 2 minus now remember I always always put the 4 AC part in Brackets so bracket 4 * a which is 1 * - 5 we're going to all divide all of that by 2 a which is two so that's going to give you x = 10 plus or minus the sare < TK of - 10^ 2 Well - 10 * - 10 would be 100 positive minus 4 * 1 is 4 * - 5 would be - 20 so that's why I'm put it in Brackets because you've got 100 here minus- 20 which will then become 120 whenever we add it up we're going to divide that by two so that equals 10 plus or minus the square Ro T of uh 100 - - 20 would be 120 / 2 at this point we might as well just separate it into the two different um as possible solution so you'd have either x = 10 + < TK 20 / 2 or we're going to get or x = 10us > 120 / 2 so we just need to work these out in the calculator and then we'll get our Solutions so we have got 10 plus the sare < TK of 120 equals and then divide by two gives us an answer to two decimal places so it's decimal places that would be uh 10.48 22 DP and if we do the other solution it's going to be 10 subtract the square < TK of 120 and then divide by two equals minus to two decimal places it would be .48 so we've got our two solutions to two decimal places