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Math Help - Simple equation help

  1. #1
    Member Jones's Avatar
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    Simple equation help

    Hi, im having problems with a simple equation that looks like this:
    \sqrt{x+3} = \sqrt{x-2} + \sqrt{x-5}

    After squaring both sides i have: x+3=2x-7 +2\sqrt{x-2}\cdot \sqrt{x-5} \rightarrow x-10+2 \cdot (x-2)^\frac{1}{2} \cdot (x-5) ^\frac {1}{2} Wich gives: x-10+2x^{2}-14x+20=0

    However, the anser should be 6 and i get 4. Im not sure about what i am doing wrong =/
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  2. #2
    Junior Member Serena's Girl's Avatar
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    When you multiplied sqrt(x - 2) and sqrt(x - 5) to each other, you forgot to include the square root sign in the product.

    You ended up with a final expression that doesn't include any square root signs so it became inaccurate.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Jones View Post
    Hi, im having problems with a simple equation that looks like this:
    \sqrt{x+3} = \sqrt{x-2} + \sqrt{x-5}

    After squaring both sides i have: x+3=2x-7 +2\sqrt{x-2}\cdot \sqrt{x-5} \rightarrow x-10+2 \cdot (x-2)^\frac{1}{2} \cdot (x-5) ^\frac {1}{2} Wich gives: x-10+2x^{2}-14x+20=0

    However, the anser should be 6 and i get 4. Im not sure about what i am doing wrong =/
    Starting from this step:

    x-10+2 \cdot (x-2)^\frac{1}{2} \cdot (x-5) ^\frac {1}{2}=0

    This becomes: \sqrt{x^2-7x+10}=\tfrac{1}{2}(10-x)

    Now square both sides:

    x^2-7x+10=\tfrac{1}{4}(x^2-20x+100)

    \implies 4x^2-28x+40=x^2-20x+100

    \implies 3x^2-8x-60=0

    \implies (x-6)(3x+10)=0

    \implies \color{red}\boxed{x=6} \ \text{or} \ \color{red}\boxed{x=-\tfrac{10}{3}}

    Test values:

    \sqrt{-\tfrac{10}{3}+3} = \sqrt{-\tfrac{10}{3}-2} + \sqrt{-\tfrac{10}{3}-5}

    \sqrt{\tfrac{1}{3}}i=4\sqrt{\tfrac{1}{3}}i+5\sqrt{  \tfrac{1}{3}}i

    This doesn't work...

    \sqrt{6+3} = \sqrt{6-2} + \sqrt{6-5}

    \implies 3=2+1 \implies 3=3

    x=6 is the only solution.
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  4. #4
    Member Jones's Avatar
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    Quote Originally Posted by Serena's Girl View Post
    When you multiplied sqrt(x - 2) and sqrt(x - 5) to each other, you forgot to include the square root sign in the product.

    You ended up with a final expression that doesn't include any square root signs so it became inaccurate.
    Since we have \sqrt(x-2) \cdot \sqrt(x-5)
    Its the same thing as (x-2)^\frac{1}{2} \cdot (x-5)^\frac{1}{2}

    Accordign to the exponetial rules A^b \cdot B^a = AB^{a + b} and \frac{1}{2} + \frac {1}{2} = 1

    Therefore  ((x-2) \cdot (x-5))^{1}
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Jones View Post
    Since we have \sqrt(x-2) \cdot \sqrt(x-5)
    Its the same thing as (x-2)^\frac{1}{2} \cdot (x-5)^\frac{1}{2}

    Accordign to the exponetial rules A^b \cdot B^a = AB^{a + b} and \frac{1}{2} + \frac {1}{2} = 1

    Therefore  ((x-2) \cdot (x-5))^{1}
    But you can only do that if the exponential functions have the same base.

    In the case of square roots, \sqrt{a}\cdot\sqrt{b}=\sqrt{ab} \ \forall a\geq 0, \ b\geq 0

    --Chris
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  6. #6
    Member Jones's Avatar
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    Oh, right.

    Thank you
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