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Thread: No Real Solutions

  1. #1
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    Cool No Real Solutions

    Hey,

    I came upon a problem where I cannot understand why $\displaystyle {x^2} + 6$ has no real solutions for the polynomial equation $\displaystyle f(x) = - 3({x^2} + 6){({x^2} + 4)^2}$ . Any help is greatly appreciated.

    Sam
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: No Real Solutions

    You're talking about the equation, but what's $\displaystyle f(x)$? Is $\displaystyle f(x)=0$?
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  3. #3
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    Re: No Real Solutions

    Yes, $\displaystyle f(x) = 0$:

    $\displaystyle \begin{array}{l} f(x) = - 3({x^2} + 6){({x^2} + 4)^2}\\ f(x) = 0 = - 3({x^2} + 6){({x^2} + 4)^2}\\ 0 \ne - 3\\ 0 = {x^2} + 6\\ 0 = {({x^2} + 4)^2} \end{array}$
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  4. #4
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    e^(i*pi)'s Avatar
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    Re: No Real Solutions

    Quote Originally Posted by ArcherSam View Post
    Hey,

    I came upon a problem where I cannot understand why $\displaystyle {x^2} + 6$ has no real solutions for the polynomial equation $\displaystyle f(x) = - 3({x^2} + 6){({x^2} + 4)^2}$ . Any help is greatly appreciated.

    Sam
    Quote Originally Posted by ArcherSam View Post
    Yes, $\displaystyle f(x) = 0$:

    $\displaystyle \begin{array}{l} f(x) = - 3({x^2} + 6){({x^2} + 6)^2}\\ f(x) = 0 = - 3({x^2} + 6){({x^2} + 6)^2}\\ 0 \ne - 3\\ 0 = {x^2} + 6\\ 0 = {({x^2} + 6)^2} \end{array}$
    You've changed from $\displaystyle (x^2+4)^2$ in post 1 to $\displaystyle (x^2+6)^2$ in post 3. Which is correct?

    In the case of post 3 write $\displaystyle f(x) = -3(x^2+6)(x^2+6)^2 = -3(x^2+6)^3$
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  5. #5
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    Re: No Real Solutions

    Quote Originally Posted by e^(i*pi) View Post
    You've changed from $\displaystyle (x^2+4)^2$ in post 1 to $\displaystyle (x^2+6)^2$ in post 3. Which is correct?

    In the case of post 3 write $\displaystyle f(x) = -3(x^2+6)(x^2+6)^2 = -3(x^2+6)^3$
    $\displaystyle {({x^2} + 4)^2}$ is correct. Sorry about the error.
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  6. #6
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    Re: No Real Solutions

    suppose that:

    $\displaystyle x^6 + 6 = 0\ \implies x^2 = -6$

    which real numbers have negative squares?

    $\displaystyle x = 0 \implies x^2 = 0,\ x > 0 \implies x^2 > 0$

    $\displaystyle x < 0 \implies -x > 0,\ x^2 = (1)(x^2) = (-1)^2x^2 = (-x)^2 > 0$
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  7. #7
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    Re: No Real Solutions

    LOL, the non-negative property of even exponents has conquered me again. Thanks Deveno.
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  8. #8
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    Re: No Real Solutions

    When findin the zeros for. $\displaystyle x^2+6 = 0$
    $\displaystyle x^2 = -6$
    $\displaystyle x^2 = {sqrt}{-6}$

    *Not possible to have square root of a negative number, hence
    no real roots. Also if you graph the parabola for the above
    Quad equation, you will see it is a transaltion of 6units vertically
    up the yaxis, and the graph doesn't cut the xaxis; hence no solutions.
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  9. #9
    Member BobBali's Avatar
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    Re: No Real Solutions

    When findin the zeros for. $\displaystyle x^2+6 = 0$
    $\displaystyle x^2 = -6$
    $\displaystyle x = {sqrt}{-6}$

    *Not possible to have square root of a negative number, hence
    no real roots. Also if you graph the parabola for the above
    Quad equation, you will see it is a transaltion of 6units vertically
    up the yaxis, and the graph doesn't cut the xaxis; hence no solutions.
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