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Thread: Quadratic in form

  1. #1
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    Quadratic in form

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

    $\displaystyle Let A=x^{2}$

    $\displaystyle A^2+A-42=0$

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

    $\displaystyle x^{2}=-7$ $\displaystyle x^{2}=6$

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

    I need help I don't think I did this correctly.
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  2. #2
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    Re: Quadratic in form

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

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

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

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

    When you have to solve it in $\displaystyle \mathbb{C}$ then:
    $\displaystyle x= \pm i\sqrt{13}$ 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 $\displaystyle A^2+6A-91=0$ 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 $\displaystyle a^2+6a-91=0$ can that be done?
    $\displaystyle (a+3)^2=100$
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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 $\displaystyle A+3=\pm 10$ or $\displaystyle A=-13\text{ or }A=7$
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