factor to solve the equation

**tonie** · March 20th 2011, 06:58 AM

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

**skeeter** · March 20th 2011, 07:10 AM

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 $r$

**e^(i*pi)** · March 20th 2011, 07:12 AM

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

**tonie** · March 20th 2011, 07:35 AM

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!

**Prove It** · March 20th 2011, 07:05 PM

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

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

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

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

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

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

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

**HallsofIvy** · March 21st 2011, 04:28 AM

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, $r= 3- 1= 2$ .

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