# Math Help - factor to solve the equation

1. ## factor to solve the equation

r^2 + 6r + -8 = 0
-8 -8
= r^2 + 6r = -8 + 9
(r+3)^2 = 1
I am stuck and the book has my answers to be -2 and -4

2. Originally Posted by tonie
r^2 + 6r + -8 = 0
-8 -8
= r^2 + 6r = -8 + 9
(r+3)^2 = 1
I am stuck and the book has my answers to be -2 and -4
I have no idea what method you are using to solve this quadratic equation. If the solutions are -2 and -4 , then this should be the original equation ...

$r^2 + 6r + 8 = 0$

factor the left side ...

$(r + 2)(r + 4) = 0$

setting each factor equal to 0 will yield the given solutions for $r$

3. You're completing the square rather than factoring. Factoring is finding out which values of $a$ and $b$ satisfy the equation $(r-a)(r-b) = r^2+6r+8$
Your syntax is confusing, I have assumed the constant term is positive (+8) because that factors.

What two numbers multiply to give +8 and sum to +6

4. While thank you so much, you fully understand my issue even though my syntax was confusing. That is exactly what the book had for the problem and answer!

5. If you wish to factorise by completing the square, and assuming that your originally equation is $\displaystyle r^2 + 6r + 8 = 0$...

$\displaystyle r^2 + 6r + 8 = 0$

$\displaystyle r^2 + 6r + 3^2 - 3^2 + 8 = 0$

$\displaystyle (r + 3)^2 - 1 = 0$

$\displaystyle (r + 3)^2 - 1^2 = 0$

$\displaystyle (r + 3 - 1)(r + 3 + 1) = 0$

$\displaystyle (r + 2)(r + 4) = 0$.

6. Or, after you have $(r+ 3)^2= 1$, as in your first post, just take the square root of both sides:
[math\]r+ 3= \pm 1[/tex] so that $r= 3\pm 1$.

Taking the "+" sign, $r= 3+ 1= 4$ and taking the "-" sign, $r= 3- 1= 2$.