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Math Help - Roots

  1. #1
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    Roots

    Show that (4-x)\cdot\sqrt{2+a}\cdot\sqrt[3]{a}\cdot\sqrt[6]{3a+4}=4 if a=1+\sqrt5.
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  2. #2
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    Quote Originally Posted by james_bond View Post
    Show that (4-{\color{red}a})\cdot\sqrt{2+a}\cdot\sqrt[3]{a}\cdot\sqrt[6]{3a+4}=4 if a=1+\sqrt5.
    If a=1+\sqrt5 then a^2 = 6+2\sqrt5 and 3a+4 = 7+3\sqrt5. Multiply these together to check that a^2(3a+4) = 72+32\sqrt5.

    Next, (2+a)^3 = (3+\sqrt5)^3. check that is also equal to 72+32\sqrt5. Therefore \sqrt{2+a} = \bigl(a^2(3a+4)\bigr)^{1/6} = \sqrt[3]{a}\cdot\sqrt[6]{3a+4}.

    Therefore (4-a)\cdot\sqrt{2+a}\cdot\sqrt[3]{a}\cdot\sqrt[6]{3a+4} = (4-a)(2+a) = (3-\sqrt5)(3+\sqrt5) = 9-5=4.
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