How do you solve #x^2 + 18 = –9x#?

2 Answers
Aug 21, 2015

Answer:

#x_(1,2) = (-9 +- 3)/2#

Explanation:

Start by getting all the terms on one side of the equation to get it classic quadratic form

#x^2 + 9x + 18 = 0#

Now you can use the quadratic formula to help you find the two roots. For a general form quadratic equation

#color(blue)(ax^2 + bx + c = 0)#

the quadratic formula is

#color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a))#

In your case, you have #a = 1#, #b = 9#, and #c=18#. This means that you have

#x_(1,2) = (-9 +- sqrt(9^2 - 4 * 1 * 18))/(2 * 1)#

#x_(1,2) = (-9 +- sqrt(9))/2#

#x_(1,2) = (-9 +- 3)/2 = {(x_1 = (-9 -3)/2 = color(green)(-6)), (x_2 = (-9 + 3)/2 = color(green)(-3)) :}#

Aug 22, 2015

Answer:

Solve #y = x^2 + 9x + 18 = 0#

Ans: -3 and -6

Explanation:

I use the new Transforming Method (Google, Yahoo Search)
Find the 2 real roots, knowing sum (-b = -9) and product (c = 18).
Both roots are negative (Rule of Signs)
Factor pairs of (18) --> (-2, -9)(-3, -6). This sum is -9 = -b. Then, the 2 real roots are -3 and -6.