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Math Help - Pulling out X to simplify, is this a correct method?

  1. #1
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    Pulling out X to simplify, is this a correct method?

    Hey guys, I'm sorry for the extremely simple question, but I'm doing math by Distance Learning, and my tutor seems to be MIA. =(

    So I've got the equation f(x)=x^2-4\sqrt{x} and I need to find my x intercept, which i'm doing by setting f(x)=0.

    This gives me

    0=x^2-4\sqrt{x}

    Now I'm trying to simplify it to isolate x by pulling out an x, but I'm not sure if this is a valid rule or not.

    0=x^2-4\sqrt{x}
    0=x^2-4x^{1/2}
    0=x(x-4(1)^{-1/2})

    If I can do this, I think that next I would have 0=x(x-4) which would mean that x cannot be equal to 0 or 4, and I think that these would be my x-intercept(s)?

    Thoughts? I don't know why this is spiraling my brain out of control.

    Also, in the original function, to find the y intercept, i plug in 0 for x right? which would give me 0, meaning that the y-intercept is through the origin? Can I use this to decide that the x-intercept is also 0?

    Thanks in advance for ANY help!
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  2. #2
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    Quote Originally Posted by Kasper View Post
    Hey guys, I'm sorry for the extremely simple question, but I'm doing math by Distance Learning, and my tutor seems to be MIA. =(

    So I've got the equation f(x)=x^2-4\sqrt{x} and I need to find my x intercept, which i'm doing by setting f(x)=0.

    This gives me

    0=x^2-4\sqrt{x}

    Now I'm trying to simplify it to isolate x by pulling out an x, but I'm not sure if this is a valid rule or not.

    0=x^2-4\sqrt{x}
    0=x^2-4x^{1/2}
    0=x(x-4(1)^{-1/2})

    If I can do this, I think that next I would have 0=x(x-4) which would mean that x cannot be equal to 0 or 4, and I think that these would be my x-intercept(s)?

    Thoughts? I don't know why this is spiraling my brain out of control.

    Also, in the original function, to find the y intercept, i plug in 0 for x right? which would give me 0, meaning that the y-intercept is through the origin? Can I use this to decide that the x-intercept is also 0?

    Thanks in advance for ANY help!
    That's the first time I've seen that tried.
    And this comment does not help.

    The only reason I commented -- it is the easiest way to get back to this question.
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by Kasper View Post
    Hey guys, I'm sorry for the extremely simple question, but I'm doing math by Distance Learning, and my tutor seems to be MIA. =(

    So I've got the equation f(x)=x^2-4\sqrt{x} and I need to find my x intercept, which i'm doing by setting f(x)=0.

    This gives me

    0=x^2-4\sqrt{x}

    Now I'm trying to simplify it to isolate x by pulling out an x, but I'm not sure if this is a valid rule or not.

    0=x^2-4\sqrt{x}
    0=x^2-4x^{1/2}
    0=x(x-4(1)^{-1/2})

    If I can do this, I think that next I would have 0=x(x-4) which would mean that x cannot be equal to 0 or 4, and I think that these would be my x-intercept(s)?

    Thoughts? I don't know why this is spiraling my brain out of control.

    Also, in the original function, to find the y intercept, i plug in 0 for x right? which would give me 0, meaning that the y-intercept is through the origin? Can I use this to decide that the x-intercept is also 0?

    Thanks in advance for ANY help!
    Hi Kasper,

    0=x^2-4\sqrt{x}
    0=x^2-4x^{1/2}
    0=x(x-4(1)^{-1/2})
    x^{1/2} is the common factor, not x.

    x^{1/2}(x^{3/2}-4)=0

    \sqrt{x}=0 \ \ or \ \ x^{3/2}-4=0

    {\color{red}x=0} \ \ or \ \ \sqrt{x^3}-4=0

    \sqrt{x^3}=4

    x^3=16

    {\color{red}x=\sqrt[3]{16} \ \ or \ \ x=2\sqrt[3]{2}}

    But, I believe I'd do it another way.

    x^2-4\sqrt{x}=0

    x^2=4\sqrt{x}

    x^4=16x

    x^4-16x=0

    x(x^3-16)=0

    {\color{red}x=0} \ \ or \ \ x^3=16

    {\color{red}x=\sqrt[3]{16}\ \ or \ \ x=2\sqrt[3]{2}}

    In either case, your x-intercepts are {\color{red}0} \ \ and \ \ {\color{red}2\sqrt[3]{2}}

    Your y-intercept is indeed 0.
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