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  1. #1
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    Question 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
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  2. #2
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    Quote Originally Posted by tonie View Post
    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
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  3. #3
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    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
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  4. #4
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    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!
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  5. #5
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    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.
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  6. #6
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    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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