Quadratic equation part 2

welcome to part two of the presentation on quadratic equations well I think I thoroughly confuse you the last time around so let me see if I can fix that a bit by doing several more examples so let's just start with a review of what the quadratic equation is the quadratic equation says if I'm trying to solve the equation a x squared plus BX plus C equals 0 then the solution or the solutions because there's usually two times that it intersects the x axis or two solutions for this equation is x equals minus b plus or minus the square root of b squared minus 4 times a times C and all of that over 2a so let's do a problem and hopefully this should make a little more sense that's a 2 on the bottom so let's say I had the equation minus 9 x squared minus 9 X plus 6 equals 0 so in this example what's a well a is the coefficient on the x squared term the x squared term is here the coefficient is minus 9 so let's write that a equals minus 9 what does B equal B is the coefficient on the X term so that's this term here so B is also equal to minus 9 and C is a constant term which is in this example is 6 so C is equal to 6 now we just substitute these values into the actual quadratic equation so negative B so it's negative times negative 9 right that's B plus or minus the square root or B squared so that's 81 right negative 9 squared minus 4 times negative nine right that's a times C which is six and all of that over 2 times negative 9 which is minus 18 all right 2 times negative 2 a 2 times negative 9 let's try to simplify this up here well negative negative 9 that's positive 9 plus or minus the square root of 81 C 4 this is negative 4 times negative 9 negative 4 times negative 9 is positive 36 and then positive 36 times 6 is let's see 30 times 6 is 180 and then 180 plus another 36 is 2 16 plus 216 is that right 180 plus 36 right is 216 all of that over 2a 2a we already said is minus 18 so you simplify that more that's 9 plus or minus the square root 81 plus 216 that's 80 plus 217 that's 297 297 and all of that over minus 18 now this is actually the hardest part with the quadratic equation is actually oftentimes just simplifying this expression we have to figure out if we can simplify this radical let's see one way to just figure out if a number is divisible by 9 is to actually add up the digits and see if the digits are divisible 9 then in this case it is 2 plus 9 plus 7 is equal to 18 so let's see how many times 9 goes into that I'll do it on the side here I don't want to be too messy 9 goes into 297 it goes it goes the 3 3 times 27 27 goes 33 times right so this is the same thing as 9 plus or minus the square root of 9 times 33 / - 18 and 9 is a perfect square that's why I actually wanted to see if 9 would work because that's the only way I could get it out of the radical if it's a perfect square as you learned that exponent rules number one module so this is equal to 9 plus or minus 3 times the square root of 33 and all of that over - 18 we're almost done we can actually simplify it because 9 3 and minus 18 are all divisible by 3 so this is equal to let's divide everything by 3 3 plus or minus plus or minus the square root of 33 over minus 6 and we are done so as you see the hardest thing with the quadratic equation is often just simplifying the expression but what we've said I know you might have lost track on we did all of this math is we said this equation minus 9x squared minus 9x plus 6 now we found two x-values that would satisfy this equation and make make it equal to 0 one x value is x equals 3 plus the square root of 33 over minus 6 and the second value is 3 minus the square root of 33 over minus 6 and you might want to think about why + where we have that plus or minus we'll add that plus or minus because the square root could actually be a positive or a negative number let's do another problem hopefully this one will be a little bit a little bit simpler let's say I wanted to solve minus 8x squared plus 5x plus 9 now I'm going to assume that you've memorized a quadratic equation because that's something you should do or you should write down on a piece of paper but the quadric issue is negative B so B is 5 right so it's we're trying to solve that equals 0 so negative B so negative 5 plus or minus the square root of b-squared that's 5 squared 25 minus 4 times a which is minus 8 times C which is 9 and all of that over 2 times a well a is minus 8 so that's all of that is over a minus 16 so let's simplify this expression up here well that's equal to minus 5 plus or minus the square root of 25 let's see 4 times 8 is 32 and the negatives cancel out so that's positive 32 times 9 positive 32 times 9 let's see 30 times 9 is 270 its 288 288 I think right 3232 70 and then we have it 288 we have all of that over minus 16 now simplify it more minus 5 plus or minus the square root 25 plus 288 is 313 I believe 313 all of that over minus 16 and I think I'm not 100% sure although I'm pretty sure I haven't checked it the 313 can't be factored into a product of a perfect square and another number in fact it actually might be a prime number that's something that you might want to check out so if that is the case and we've gotten it in completely simplified form and we say there are two solutions to X values that will make this equation true one of them is X is equal to minus 5 plus the square root of 313 over minus 16 and the other one is X is equal to minus 5 minus the square root of 313 over minus 60 hopefully those two examples will give you a good sense of how to use the quadratic equation I might add some more modules and then once you master this I'll actually teach you how to solve quadratic equations when you actually get a negative number under the radical very interesting anyway I hope you can do the module now and maybe I'll add a few more presentations because this isn't the easiest module but I hope you have fun bye