Solving the equation x^2+7x=18?

Question

How do you solve the equation x²+7x=18?

Thank you!

Answer 1

Subtract 18 from both sides to get the equation set to 0:

x² + 7x - 18 = 0

Now factor. Given the negative sign in front of the 18, you will have (x +   ) (x -     ) as your factors, with the larger one being positive. Now, which factors of 18 can be combined to get a result of 7? I'll leave that for you. 

Answer 2

x²+7x-18=0  is an eqaution of a quadratic form ax²+bx +c = 0

For the above equation a=1, b= 7,  and c = -18

to test how many roots are for this equation compute b² - 4 ac, (7)²-4(1)(-18) = 121

Since b2- 4ac > 0 or positive the equation has two roots, please note that if b2-4ac < 0 or negative, the equation has no roots, and if b2-4ac = 0, the equation has one root and therefore one solution.

The root  are found by   x = (- b ± √b²-4ac ) / 2a , and x = (-7  ± √121 ) / 2 (1) = ( -7 ± 11) / 2

The two roots are x1 = ( -7 - 11) /2 = -18 /2 = - 9

and                     x2 = ( -7 + 11) / 2 = 4 /2 =  2

To check you answer substitute each x value in the equation to get zero

for x1 =  -9 , then (-9)²+7(-9)-18 = 81 -63 -18 = 0

for x2 = 2, then (2)²+ 7 (2) -18 = 4 + 14 - 18 = 0

Threfore both values of x are valid for the equations, and the equation has two roots x = -9 , and x = 2