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Thread: Algebraic manipulations and solutions

  1. #1
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    Question Algebraic manipulations and solutions

    Hi,

    Can someone help clarify why the solution to this expression does not result in both solutions equating to 2 when substituted in for x? I'm referring to when x = 4, since (4) + (4)^(1/2) does not equal 2. What is going on here?

    Please let me know!
    - otownsend
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  2. #2
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    Re: Algebraic manipulations and solutions

    What expression are you talking about?
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    Re: Algebraic manipulations and solutions

    Sorry, I forgot to include it. The expression is x + (x)^(1/2) = 4
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  4. #4
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    Re: Algebraic manipulations and solutions

    Quote Originally Posted by otownsend View Post
    The expression is x + (x)^(1/2) = 4
    That's an equation, not an expression...
    It simplifies to x^2 - 9x + 16 = 0

    Now what's your question?
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  5. #5
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    Re: Algebraic manipulations and solutions

    Quote Originally Posted by DenisB View Post
    That's an equation, not an expression...
    It simplifies to x^2 - 9x + 16 = 0

    Now what's your question?
    @otownsend, just so you can see how to obtain that simplified equation, here are the steps:

    $x+x^{1/2} = 4$

    Subtract x from both sides:

    $x^{1/2} = 4-x$

    Square both sides:

    $x = (4-x)^2 = x^2-8x+16$

    Subtract x from both sides:

    $x^2-9x+16=0$
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  6. #6
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    Re: Algebraic manipulations and solutions

    my question is why is it that the solutions for the equation, which are x = 1 and x = 4, do not both work when substituting them into the original equation before algebraic manipulation? x = 1 works, but x = 4 does not work. I forget why this is the case and would appreciate someone to explain this.

    Also I am referring to the equation x + (x)^(1/2) = 2. I mentioned x + (x)^1/2 = 4 in my previous post, but that was a mistake. Please take this into account.
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  7. #7
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    Re: Algebraic manipulations and solutions

    Quote Originally Posted by otownsend View Post
    my question is why is it that the solutions for the equation, which are x = 1 and x = 4, do not both work when substituting them into the original equation before algebraic manipulation? x = 1 works, but x = 4 does not work. I forget why this is the case and would appreciate someone to explain this.

    Also I am referring to the equation x + (x)^(1/2) = 2. I mentioned x + (x)^1/2 = 4 in my previous post, but that was a mistake. Please take this into account.
    Then, $x^{1/2} = 2-x$

    When we square both sides, we introduce another solution. When we square both sides, we obtain $x = (2-x)^2$. If we were to go backwards and take the square root of both sides, we could get $\pm \sqrt{x} = 2-x$. But, the original problem was only $\sqrt{x} = 2-x$. So, by squaring, we actually added another possible solution. You find the correct solution(s) by plugging the possible solutions into the original equation to determine which one(s) is/are correct.
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  8. #8
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    Re: Algebraic manipulations and solutions

    Gotcha! Thanks a lot
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