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Math Help - Quadratic Formula Problem

  1. #1
    Newbie fetuslasvegas's Avatar
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    Quadratic Formula Problem

    Can somebody help break this problem down for me using the quadratic formula? Thanks!

    x^2+2x-2=0

    I know the answer is x=-1 plusminus square root 3 (don't know how to write that on here)
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by fetuslasvegas View Post
    Can somebody help break this problem down for me using the quadratic formula? Thanks!

    x^2+2x-2=0

    I know the answer is x=-1 plusminus square root 3 (don't know how to write that on here)
    For ax^2+bx+c = 0 then x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

    If you compare your equation to ax^2+bx+c=0 then you'll see that a=1, b=2 and c=-2. The place most people split up is a sign error when calculating b^2-4ac
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  3. #3
    Newbie fetuslasvegas's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    For ax^2+bx+c = 0 then x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

    If you compare your equation to ax^2+bx+c=0 then you'll see that a=1, b=2 and c=-2. The place most people split up is a sign error when calculating b^2-4ac
    So I know this is sort of what it would look like when I go through the process, Where do I go from there to get the answer?

    x = \frac{-2 \pm \sqrt{2^2-4*1*-2}}{2*1}
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by fetuslasvegas View Post
    So I know this is sort of what it would look like when I go through the process, Where do I go from there to get the answer?

    x = \frac{-2 \pm \sqrt{2^2-4*1*-2}}{2*1}
    x = \frac{-2 \pm \sqrt{4+8}}{2}

    x = \frac{-2 \pm \sqrt{12}}{2}

    x = \frac{-2 \pm 2\sqrt{3}}{2}

    so,

    x = \frac{-2 + 2\sqrt{3}}{2} = {\sqrt{3} -1}
    OR
    x = \frac{-2 - 2\sqrt{3}}{2}= {-\sqrt{3} -1}

    Both values of x satisfy your equation
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  5. #5
    Newbie fetuslasvegas's Avatar
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    Quote Originally Posted by harish21 View Post
    x = \frac{-2 \pm \sqrt{4+8}}{2}

    x = \frac{-2 \pm \sqrt{12}}{2}

    x = \frac{-2 \pm 2\sqrt{3}}{2}

    so,

    x = \frac{-2 + 2\sqrt{3}}{2} = {\sqrt{3} -1}
    OR
    x = \frac{-2 - 2\sqrt{3}}{2}= {-\sqrt{3} -1}

    Both values of x satisfy your equation


    Ok I get mostly everything except where the -1 comes in?


    The way I see the answer is x = \frac{\sqrt{3}}{2}
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  6. #6
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by fetuslasvegas View Post
    Ok I get mostly everything except where the -1 comes in?


    The way I see the answer is x = \frac{\sqrt{3}}{2}
    x = \frac{-2 + 2\sqrt{3}}{2} =  \frac{2(-1 + \sqrt{3})}{2}= -1 + \sqrt{3}= \sqrt{3}-1

    same for the second part
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  7. #7
    Newbie fetuslasvegas's Avatar
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    Quote Originally Posted by harish21 View Post
    x = \frac{-2 + 2\sqrt{3}}{2} =  \frac{2(-1 + \sqrt{3})}{2}= -1 + \sqrt{3}= \sqrt{3}-1

    same for the second part
    I see how it makes sense, I just am not getting where this -1 is coming from or what you are doing to get the -1. Is it coming from moving the negative sign from the first two which for some reason subtracts from the second 2?
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  8. #8
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    Quote Originally Posted by fetuslasvegas View Post
    I see how it makes sense, I just am not getting where this -1 is coming from or what you are doing to get the -1. Is it coming from moving the negative sign from the first two which for some reason subtracts from the second 2?
    -2+2\sqrt{3} = 2\times -1+2\times \sqrt{3}

    Now 2 is common to both terms so it can be factored out giving

     2\times -1+2\times \sqrt{3} = 2(-1+\sqrt{3})

    If you cannot follow this please revise basic factorisation before asking additional questions.
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  9. #9
    Newbie fetuslasvegas's Avatar
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    Quote Originally Posted by pickslides View Post
    -2+2\sqrt{3} = 2\times -1+2\times \sqrt{3}

    Now 2 is common to both terms so it can be factored out giving

     2\times -1+2\times \sqrt{3} = 2(-1+\sqrt{3})

    If you cannot follow this please revise basic factorisation before asking additional questions.

    Ok I'm still not understanding, I apologize, I'll continue to try to figure it out from my book.
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