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Math Help - Solving for the square roots of binomials

  1. #1
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    Solving for the square roots of binomials

    I have no idea how to do this one:

    Sqrt(3x+4) + Sqrt(x-3) = 5

    I did guess and check and found that x = 4, but I want a more elegant solution. I did try squaring both sides and came up with

    3x + 4 + x - 3 + 2Sqrt(3x^2 - 5x -12) = 25 but, long story short, that did not turn out right for me. Am i on the right track?
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  2. #2
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    Quote Originally Posted by satx View Post
    I have no idea how to do this one:

    Sqrt(3x+4) + Sqrt(x-3) = 5

    I did guess and check and found that x = 4, but I want a more elegant solution. I did try squaring both sides and came up with

    3x + 4 + x - 3 + 2Sqrt(3x^2 - 5x -12) = 25 but, long story short, that did not turn out right for me. Am i on the right track?
    HI

    Yes , you will need to start by squaring .

    [\sqrt{3x+4}+\sqrt{x-3}]^2=[5]^2

    4x+1+2\sqrt{3x^2-5x-12}=25

    12-2x=\sqrt{3x^2-5x-12}

    Then square both sides once again and simplify .

    x^2-43x+156=0

    (x-4)(x-39)=0

    x=4 , 39

    Now test these 2 values . Only 4 is valid .
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  3. #3
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    cheers mate
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  4. #4
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    Hello, satx!

    Another approach . . .


    \sqrt{3x+4} + \sqrt{x-3} \:=\: 5
    Isolate a radical (get a radical "by itself"):

    . . \sqrt{3x+4} \;=\;5-\sqrt{x-3}


    Square both sides: . (\sqrt{3x+4})^2 \;=\;(5-\sqrt{x-3})^2 \quad\Rightarrow\quad 3x + 4 \;=\;25 - 10\sqrt{x-3} + x-3

    . . 10\sqrt{x-3} \;=\;18 - 2x \quad\Rightarrow\quad 5\sqrt{x-3} \;=\;9-x


    Square both sides: . (5\sqrt{x-3})^2 \;=\;(9-x)^2 \quad\Rightarrow\quad 25(x-3) \;=\;81 - 18x + x^2

    . . x^2 - 43x + 156 \:=\:0 \quad\Rightarrow\quad (x-4)(x-39) \:=\:0 \quad\rightarrow\quad x \:=\:4,\:39


    As mathaddict pointed out, x = 39 is extraneous.

    . . Therefore, only x = 4 is a solution.

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