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Math Help - Proving equations with radicals.

  1. #1
    Junior Member Kaloda's Avatar
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    Proving equations with radicals.

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  2. #2
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    Re: Proving equations with radicals.

    We will need two formulas:
    x^3-y^3=(x-y)(x^2+xy+y^2)
    and
    (x-y)^2=x^2-2xy+y^2

    Let \sqrt[12]x=a

    Then we are given that a^3- \frac{1}{a^3}=14

    We need to show that a^2+\frac{1}{a^2}=6.

    Expand what we are given by first formula: a^3- \frac{1}{a^3}=\left(a-\frac 1a\right)\left(a^2+\frac aa+\frac {1}{a^2}\right)=14
    or
    \left(a-\frac 1a\right)\left(a^2+\frac {1}{a^2}+1\right)=14

    Since \left(a-\frac1a\right)^2=a^2-2 \frac aa+\frac{1}{a^2}=a^2+\frac{1}{a^2}-2 then \left(a-\frac1a\right)^2+2=a^2+\frac{1}{a^2}
    Now, equation can be rewritten as
    \left(a-\frac 1a\right)\left(\left(a-\frac1a\right)^2+2+1\right)=14
    Or
    \left(a-\frac 1a\right)\left(\left(a-\frac1a\right)^2+3\right)=14
    For simplicity, make substitution t=a-\frac 1a

    Then we have that t(t^2+3)=14
    There is only one real solution t=2, so a-\frac 1a=2
    Finally, a^2+\frac{1}{a^2}=\left(a-\frac1a\right)^2+2=2^2+2=6
    That's all.
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  3. #3
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    Re: Proving equations with radicals.

    You're given


    Let \sqrt[6]{x} + \frac{1}{\sqrt[6]{x}} = k. Cubing both sides of this equation yields


    \sqrt[6]{x}^3 + 3 \sqrt[6]{x}^2(\frac{1}{\sqrt[6]{x}}) + 3\sqrt[6]{x}(\frac{1}{\sqrt[6]{x}})^2 + (\frac{1}{\sqrt[6]{x}})^3 = k^3


    \sqrt{x} + \frac{1}{\sqrt{x}} + 3(\sqrt[6]{x} + \frac{1}{\sqrt[6]{x}}) = k^3, substitute \sqrt[6]{x} + \frac{1}{\sqrt[6]{x}} with k.


    \sqrt{x} + \frac{1}{\sqrt{x}} + 3k = k^3. Here, it helps to know that \sqrt{x} + \frac{1}{\sqrt{x}} is. However, we know that \sqrt[4]{x} - \frac{1}{\sqrt[4]{x}} = 14. Squaring both sides of this equation yields

    \sqrt{x} + \frac{1}{\sqrt{x}} - 2 = 14^2 = 196 \Rightarrow \sqrt{x} + \frac{1}{\sqrt{x}} = 198. Substitute into the previous equation to obtain


    198 + 3k = k^3. Here it is evident that k = 6 is a root (the other two roots are non-real). Hence we are done.
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  4. #4
    Junior Member Kaloda's Avatar
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    Re: Proving equations with radicals.

    Thank you VERY VERY MUCH!
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  5. #5
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    Re: Proving equations with radicals.

    Quote Originally Posted by Kaloda View Post
    Click image for larger version. 

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    Let x^.25 + x^-.25 =k
    x^.25-x-.25=14 Adding 2x^.25=14+k Subtracting 2x^-.25=k-14
    So 2/(14+k)=x^-.25 Sub for x^-.25 in last equation of previous line 4/(14+k)=k-14 So k^2-196=4 k^2=200 k= square root 200
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