Assume this quadratic equation equals zero. Then :-x2 - 13x + 40 = 0, this can be factored (x - 8)(x - 5) = 0This holds true when either x - 8 = 0 or x - 5 = 0When x - 8 = 0 then x = 8 : When x - 5 = 0 then x = 5
You should look for two numbers whose product is 40, and whose sum is 13. Experimenting a little (trying different pairs of factors for 40), you can quickly realize that this works for 8 and 5. Therefore, x2 + 13x + 40 = (x + 8) (x + 5) that would be 8 and 5 If you want to use the quadratic equation to solve it instead of factoring, the equation is x = (-b + or - the square root of (b2- 4ac))/2a a in this case is 1, b is 13, and c is 40 So, plug a, b, and c into the above equation and do it once with adding and once with subtracting. There will be two x values given if you do it both times, 8 and 5.
(x - 5)(x - 8)
X3 + 13X2 + 40X factor out an X X(X2 + 13X + 40) factor the quadratic expression, what two factors of 40 add up to 13? X(X + 8)(X + 5) works the same if that is a negative in front of X3, just sign changes
A quadratic equation in the form of: x2-54x+560 = 0 whose solutions are x = 14 and x = 40
Assume this quadratic equation equals zero. Then :-x2 - 13x + 40 = 0, this can be factored (x - 8)(x - 5) = 0This holds true when either x - 8 = 0 or x - 5 = 0When x - 8 = 0 then x = 8 : When x - 5 = 0 then x = 5
You should look for two numbers whose product is 40, and whose sum is 13. Experimenting a little (trying different pairs of factors for 40), you can quickly realize that this works for 8 and 5. Therefore, x2 + 13x + 40 = (x + 8) (x + 5) that would be 8 and 5 If you want to use the quadratic equation to solve it instead of factoring, the equation is x = (-b + or - the square root of (b2- 4ac))/2a a in this case is 1, b is 13, and c is 40 So, plug a, b, and c into the above equation and do it once with adding and once with subtracting. There will be two x values given if you do it both times, 8 and 5.
X2 + 13X - 40 = 0I can't factor this, so the quadratic equation will be used.X = - b (+/-) sqrt(b2 - 4ac)/2aa = 1b = 13c = - 40X = - 13 (+/-) sqrt[(132 - 4(1)(- 40)]/2(1)X = - 13 (+/-) sqrt(169 + 160)/2X = [- 13 (+/-) sqrt(329)]/2=========================ugly exact answerInexact answers.X = ~ 2.569X = ~ - 15.569
x2 + 13x + 40 = x2 + 5x + 8x + 40 = x(x + 5) + 8(x + 5) = (x + 8)(x + 5)
-x^2 -13x -40 = -(x^2 + 13x + 40) = -(x + 5)*(x + 8)-x^2 -13x -40 = -(x^2 + 13x + 40) = -(x + 5)*(x + 8)-x^2 -13x -40 = -(x^2 + 13x + 40) = -(x + 5)*(x + 8)-x^2 -13x -40 = -(x^2 + 13x + 40) = -(x + 5)*(x + 8)
(x - 5)(x - 8)
A quadratic equation in the form of: x2-54x+560 = 0 whose solutions are x = 14 and x = 40
X3 + 13X2 + 40X factor out an X X(X2 + 13X + 40) factor the quadratic expression, what two factors of 40 add up to 13? X(X + 8)(X + 5) works the same if that is a negative in front of X3, just sign changes
You should look for two numbers whose product is 40, and whose sum is 13. Experimenting a little (trying different pairs of factors for 40), you can quickly realize that this works for 8 and 5. Therefore, x2 + 13x + 40 = (x + 8) (x + 5)
(x + 5)(x + 8)
56
This doesn't factor neatly. Applying the quadratic equation, we find two imaginary solutions: (-17 plus or minus the square root of -671) divided by 12x = -1.4166 + 2.158638974498103ix = -1.4166 - 2.158638974498103iwhere i is the square root of negative one.