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

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- Mar 20th 2011, 06:58 AMtoniefactor 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 - Mar 20th 2011, 07:10 AMskeeter
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 $\displaystyle r$ - Mar 20th 2011, 07:12 AMe^(i*pi)
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 - Mar 20th 2011, 07:35 AMtonie
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!

- Mar 20th 2011, 07:05 PMProve It
If you wish to factorise by completing the square, and assuming that your originally equation is $\displaystyle \displaystyle r^2 + 6r + 8 = 0$...

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

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

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

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

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

$\displaystyle \displaystyle (r + 2)(r + 4) = 0$. - Mar 21st 2011, 04:28 AMHallsofIvy
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, $\displaystyle r= 3- 1= 2$.