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Math Help - I can't figure out this radical equation?

  1. #1
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    I can't figure out this radical equation?

    I'm taking Pre-Calc and Trigonometry, and the class is so poorly put together I don't even get a textbook. My teacher is of no help. I can't figure out how to solve this problem, and I desperately would like to understand. Help?

    √x+2 + √3x+4 = 2

    I have tried it over and over and have gotten different answers each time. I don't know what are the proper steps to solve an equation with two radicals on one side. :<
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  2. #2
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    \sqrt{x} +2 +\sqrt{3x}+4 = 2

    Grouping like terms together

    \sqrt{x} +\sqrt{3x}+6 = 2

    \sqrt{x} +\sqrt{3x} = -4

    \sqrt{x} +\sqrt{3}\sqrt{x} = -4

    Factoring

    \sqrt{x} (1+\sqrt{3}) = -4

    \sqrt{x} = \frac{-4}{ (1+\sqrt{3}) }

    Now square both sides
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  3. #3
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    Quote Originally Posted by pickslides View Post
    \sqrt{x} +2 +\sqrt{3x}+4 = 2

    Grouping like terms together

    \sqrt{x} +\sqrt{3x}+6 = 2

    \sqrt{x} +\sqrt{3x} = -4

    \sqrt{x} +\sqrt{3}\sqrt{x} = -4

    Factoring

    \sqrt{x} (1+\sqrt{3}) = -4

    \sqrt{x} = \frac{-4}{ (1+\sqrt{3}) }

    Now square both sides
    I get that, but the +2 and +4 are both supposed to be under the square root symbols. I wasn't sure how to express that. Maybe √(x+2) + √(3x+4) = 2 ? Sorry for the confusion, and thanks so much for the help!
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  4. #4
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    Quote Originally Posted by pickslides View Post
    \sqrt{x} +\sqrt{3x} = -4
    you may notice that having this the obvious conclusion is that there's no solution here, so it leads to think that the equation was stated as bwenner says.
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  5. #5
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    \sqrt{1+x}= 1+\frac{1}{2}x- \frac{1}{8}x^2+\frac{1}{16}x^3-\dots

    Might be outside of the scope of what you are learning but you can try this series approximation?

    Using it as a 2nd order approximation could be easy to solve.
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  6. #6
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    Quote Originally Posted by bwenner View Post

    √(x+2) + √(3x+4) = 2

    note the use of parentheses to make your equation clear.
    start by squaring both sides ...

    (x+2) + 2\sqrt{x+2}\sqrt{3x+4} + (3x+4) = 4

    2\sqrt{x+2}\sqrt{3x+4} = -(4x+2)

    \sqrt{x+2}\sqrt{3x+4} = -(2x+1)

    square again ...

    (x+2)(3x+4) = 4x^2 + 4x + 1

    3x^2 + 10x + 8 = 4x^2 + 4x + 1

    0 = x^2 - 6x -7

    0 = (x - 7)(x + 1)

    x = 7 or x = -1

    be sure to check the validity of both values in the original equation to rule out extraneous solutions.
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  7. #7
    Senior Member pacman's Avatar
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    Or this might be the right form? √(x+2) + √(3x+4) = 2.

    squaring BS,

    (x + 2) + 2√(x + 2)√(3x + 4) + 3x + 4 = 4, simplify

    2√(x + 2)√(3x + 4) = 4 - 4 - 2 - x - 3x,

    2√(x + 2)√(3x + 4) = -2 -4x = -2(1 + 2x)

    √(x + 2)√(3x + 4) = -(1 + 2x), square BS,

    (x + 2)(3x + 4) = (-1 - 2x)^2,

    3x^2 + 10 x + 8 = 1 + 4x + 4x^2, combine like terms

    (4x^2 - 3x^2) + (-10x + 4x) + (1 - 8) = 0,

    x^2 - 6x - 7 = 0, factoring

    (x - 7)(x + 1) = 0, using the zero factor theorem

    x = -1 or x = 7.

    Looking for extraneous root, we substitute the value to the original equation.

    (x + 2)^(1/2) + (3x + 4)^(1/2) = 2, for x = 7

    (7 + 2)^(1/2) + (3(7) + 4)^(1/2) = 2?

    3 + 5 = 2? not the root of the original equation,

    Next, try x = -1.

    (- + 2)^(1/2) + (3(-1) + 4)^(1/2) = 2?

    1 + 1 = 2. x = -1 is the ONLY ROOT of the above equation.

    Check by graphing: surely x = -1 is a ROOT.

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  8. #8
    Senior Member pacman's Avatar
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    see the graph, x = -1 is the only solution.
    Attached Thumbnails Attached Thumbnails I can't figure out this radical equation?-aqw.gif  
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  9. #9
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    Just a note, it's easy enough to see that the function may only have one real root, seeing as it is monotonous increasing.
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  10. #10
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    I would separate the two square roots:
    \sqrt{x+2} + \sqrt{3x+4} = 2
    \sqrt{x+2}= 2- \sqrt{3x+4}

    Now square both sides
    x+ 2= 2- 2\sqrt{3x+4}+ 2x+ 4
    2\sqrt{3x+4}= x+ 4

    Squaring again,
    2(3x+4)= (x+4)^2= x^2+ 8x+ 4
    x^2+ 8x+ 4= 6x+ 8
    x^2+ 2x- 4= 0

    Can you finish from there? Remember to check any roots in the origina equation. Squaring both sides can introduce "extraneous" roots.
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  11. #11
    Senior Member pacman's Avatar
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    HallsofIvy: ". - here
    Now square both sides
    ""

    you missed to square 2.
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