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Math Help - Another confusing radicals problem

  1. #1
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    Another confusing radicals problem

    root(2x +5) + 2root(x+6) = 5
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Trentt View Post
    root(2x +5) + 2root(x+6) = 5
    Square:

    (2x+5) + 4 sqrt(2x+5) sqrt(x+6) +4 (x+6) = 25

    so:

    4 sqrt(2x+5) sqrt(x+6) = -4-6x.

    Now square again:

    16 (2x+5)(x+6) = (4+6x)^2

    or:

    32·x^2 + 272·x + 480 = 36·x^2 + 48·x + 16

    Which simplifies to:

    4·(x^2 - 56·x - 116)=0

    This is a quadratic with roots x=58, and x=-2.

    Now we have to substitute each of these back into the original equation
    to check that they are roots (as the squarings can introduce spurious
    roots).

    root(2 58 +5) + 2root(58+6) = 27,

    so 58 is not a root of the original equation.

    root(2(-2) +5) + 2root((-2)+6) = 1+2 (2)=5

    so x=-2 is a root.

    Therefore the only solution to the original equation is x=-2.

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    Which simplifies to:

    4·(x^2 - 56·x - 116)=0

    This is a quadratic with roots x=58, and x=-2.
    So what happens to the bold underlined 4?
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  4. #4
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    Hello, Trentt!

    root(2x +5) + 2·root(x+6) .= .5

    We have: .2·root(x + 6) .= .5 - root(2x + 5)

    Square both sides: .[2·root(x + 6)]˛ .= .[5 - root(2x + 5)]˛

    . . 4(x + 6) .= .10 - 10·root(2x + 5) + 2x + 5

    . . 2x + 14 .= .-10·root(2x + 5)

    Divide by 2: .x + 7.= .-5·root(2x + 5)

    Square both sides: .(x + 7)˛ .= .[-5·root(2x + 5)]˛

    . . x˛ + 14x + 49 .= .25(2x + 5)

    which simplifies to: .x˛ - 36x - 76 .= .0

    which factors: .(x + 2)(x - 38) .= .0

    and has roots: .x = -2, 38


    But only x = -2 satisfies the equation.

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