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 ...
$\displaystyle r^2 + 6r + 8 = 0$
factor the left side ...
$\displaystyle (r + 2)(r + 4) = 0$
setting each factor equal to 0 will yield the given solutions for $\displaystyle r$
You're completing the square rather than factoring. Factoring is finding out which values of $\displaystyle a$ and $\displaystyle b$ satisfy the equation $\displaystyle (r-a)(r-b) = r^2+6r+8$
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
If you wish to factorise by completing the square, and assuming that your originally equation is $\displaystyle \displaystyle r^2 + 6r + 8 = 0$...
$\displaystyle \displaystyle r^2 + 6r + 8 = 0$
$\displaystyle \displaystyle r^2 + 6r + 3^2 - 3^2 + 8 = 0$
$\displaystyle \displaystyle (r + 3)^2 - 1 = 0$
$\displaystyle \displaystyle (r + 3)^2 - 1^2 = 0$
$\displaystyle \displaystyle (r + 3 - 1)(r + 3 + 1) = 0$
$\displaystyle \displaystyle (r + 2)(r + 4) = 0$.
Or, after you have $\displaystyle (r+ 3)^2= 1$, as in your first post, just take the square root of both sides:
[math\]r+ 3= \pm 1[/tex] so that $\displaystyle r= 3\pm 1$.
Taking the "+" sign, $\displaystyle r= 3+ 1= 4$ and taking the "-" sign, $\displaystyle r= 3- 1= 2$.