How do you solve 9x^2+11x+18=-10x+8?

1 Answer
Mar 18, 2017

1) Rearrange equation so that it is in the form ax^2 + bx +c =0
2) (Learn by heart and) use the quadratic formula to solve the equation

Answers are, in this case,
x_1 = -2/3 and x_2 = -5/3

Explanation:

Ohkay, let's do this.
1) Rearrange every term on one side, say to the left.
We now have:
9x^2 + 11x + 18 +10x -8 = 0
(the sign-change occurs because the terms "crossed" the equal sign)

Now, you can simplify a bit. We have:
9x^2 + 21x +10 = 0

This is a quadratic equation of the form
ax^2 + bx + c =0 where a,b,c are real numbers (in our case, they are integers, which are also real numbers but enough side-tracking).
To solve this, you need the quadratic formula which is, in its general form:
x_(1,2) = (-b \pm sqrt(b^2 - 4ac))/(2a)
It says x_(1,2) and there is a \pm(plus or minus) sign in this equation because there can be a maximum of two (real) solutions, x_1 and x_2 which depend on whether we use the plus sign or the minus sign in that equation.

Anyway, applying this formula to our equation, we need to put a = 9, b=21, and c=10
we get the following:
x_(1,2) = (-21 \pm sqrt(21^2 - 4*9*10))/(2*9)
i.e.
x_(1,2) = (-21 \pm sqrt(441 - 360))/(18)
i.e.
x_(1,2) = (-21 \pm sqrt(81))/(18)
i.e.
x_(1,2) = (-21 \pm 9)/(18)
so
x_1 = -12/18 and x_2 = -30/18
i.e.
x_1 = -2/3 and x_2 = -5/3

Q.E.D.