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Thread: Squre root of 1 I guess

  1. #1
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    Squre root of 1 I guess

    I begin with $\displaystyle x^2+8x=-15$. Since they say to solve by completing the square, I ignore that it can be factored as is and add one to each side. Since $\displaystyle (1/2*8)^2=16$. And get $\displaystyle x^2+8x+16=1$ after moving the -15 and adding one to each side.
    First off have I done everything correctly? Keep in mind that the problem asks for me to complete the square, which I think I have done.

    Secondly is the resulting factored equation just asking me for the square root of one? $\displaystyle (x+4)^2=1$ is the same as $\displaystyle a^2=1$. Since x would be what ever number plus four squared is equal to one.


    Just from factoring out the initial equation you can tell the answers are -3 and -5. So something has too be wrong.
    Last edited by bkbowser; Nov 11th 2009 at 08:21 AM. Reason: Clarity
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  2. #2
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    When solving (rather than merely simplifying), you have to account for all possible values. So while $\displaystyle \sqrt{1}\, =\, 1$ for simplifying, you need to use $\displaystyle \sqrt{1}\, =\, \pm 1$ for solving.

    So you actually have:

    $\displaystyle x\, +\, 4\, =\, \pm 1$

    Subtract the 4 over to the right-hand side, and simplify to get the two values. They should match what you've been given.
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  3. #3
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    OK thats why, I'm just forgeting that the square root of 1 is 1 and didnt work the problem out all the way.

    Thanks
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