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Math Help - Quadratic in form

  1. #1
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    Quadratic in form

    x^{4}+1x^{2}-42=0

    Let A=x^{2}

    A^2+A-42=0

    \left ( A+7 \right )\left ( A-6 \right )=0

    x^{2}=-7  x^{2}=6

    x=\pm \sqrt{-7} x=\pm \sqrt{6}

    I need help I don't think I did this correctly.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Quadratic in form

    It's almost correct, x=\pm \sqrt{6} is a solution of the equation but you have to reject x=\pm \sqrt{-7} because \sqrt{-7} is undefined in \mathbb{R}.
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  3. #3
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    Re: Quadratic in form

    Separate Question:
    x^{4}+6x^{2}-91=0
    I did the exact same thing but ended up with
    (A+3)^2=100
    x^2=-3\pm10
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  4. #4
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    Re: Quadratic in form

    Quote Originally Posted by theloser View Post
    Separate Question:
    x^{4}+6x^{2}-91=0
    I did the exact same thing but ended up with
    (A+3)^2=100
    x^2=-3\pm10
    x^4+6x^2-91 = (x^2+13)(x^2-7) = (x + i\sqrt{13})(x - i\sqrt{13})(x + \sqrt{7})(x - \sqrt{7})
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: Quadratic in form

    That's not correct.
    x^4+6x^2-91=0
    Let x^2=t so:
    t^2+6t-91=0
    \Leftrightarrow (t-7)(t+13)=0
    \Leftrightarrow x^2=7 \ \mbox{or} \ x^2=-13
    We can reject x^2=-13 in \mathbb{R} and the solutions are now: x=\pm \sqrt{7}

    When you have to solve it in \mathbb{C} then:
    x= \pm i\sqrt{13} is also a solution.
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  6. #6
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    Re: Quadratic in form

    I see that it can factor but I hate to factor big numbers so I instead tried to complete the square on A^2+6A-91=0 can that be done?
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  7. #7
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    Re: Quadratic in form

    Quote Originally Posted by theloser View Post
    i see that it can factor but i hate to factor big numbers so i instead tried to complete the square on a^2+6a-91=0 can that be done?
    (a+3)^2=100
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  8. #8
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    Re: Quadratic in form

    But does this give me the correct answer?
    After that it would be
    A=13
    A=7
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  9. #9
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    Re: Quadratic in form

    Quote Originally Posted by theloser View Post
    But does this give me the correct answer?
    After that it would be
    A=13
    A=7
    NO INDEED!
    It gives A+3=\pm 10 or A=-13\text{ or }A=7
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