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Math Help - Find all Real solutions

  1. #1
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    Find all Real solutions

    This is just some hw question I'm stuck on

    √(x+7) = (x+1)

    and

    (3x – 2)^5/6 = 2
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by civiliam View Post
    This is just some hw question I'm stuck on

    √(x+7) = (x+1)

    and

    (3x 2)^5/6 = 2
    \sqrt[3]{x^3+7}=x+1

    Cube both sides:

    x^3+7=x^3+3x^2+3x+1

    3x^2+3x-6=0

    x^2+x-2=0

    (x+2)(x-1)

    \boxed{x=-2 \ \ or \ \ x=1}
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  3. #3
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    Hello, civiliam!

    (3x - 2)^{\frac{5}{6}} \:= \:2
    Raise both sides to the \frac{6}{5} power . . .

    . . \bigg[(3x-2)^{\frac{5}{6}}\bigg]^{\frac{6}{5}} \;=\;2^{\frac{6}{5}}

    . . . . . . 3x-2 \;=\;2\sqrt[5]{2}

    . . . . . . . . 3x \;=\;2 + 2\sqrt[5]{2}

    . . . . . . . . . x \;=\;\frac{2 + 2\sqrt[5]{2}}{3}

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  4. #4
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    thank you both I think I need to review exponents
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