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Math Help - Factoring and rationalizing

  1. #1
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    Factoring and rationalizing

    1/ sqroot(x) - 1/2 / x-4


    I first simplified by multiplying by 2sqroot(x) = 2-sqroot(x) / (x-4) (2sqroot(x)) than i was going to rationalize the numerator by multiplying by 2+ sqroot(x) but that seemed too complicated.

    Where did i go wrong?
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  2. #2
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    Re: Factoring and rationalizing

    Quote Originally Posted by skg94 View Post
    1/ sqroot(x) - 1/2 / x-4


    I first simplified by multiplying by 2sqroot(x) = 2-sqroot(x) / (x-4) (2sqroot(x)) than i was going to rationalize the numerator by multiplying by 2+ sqroot(x) but that seemed too complicated.

    Where did i go wrong?
    Why dont you type in latex or give proper brackets?? It helps a lot....
    is your question:Rationalize and factorize \frac{1}{\sqrt{x}}-\frac{1}{2(x-4)}
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  3. #3
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    Re: Factoring and rationalizing

    I dont know how to, but well i suppose its the same but its (1/rootx - 1/2 ) / (x-4)
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  4. #4
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    Re: Factoring and rationalizing

    Hello, skg94!

    \text{Rationalize: }\:\dfrac{\frac{1}{\sqrt{x}} - \frac{1}{2}}{x-4}

    I first simplified by multiplying by \tfrac{2\sqrt{x}}{2\sqrt{x}} .and got: . \frac{2-\sqrt{x}}{2\sqrt{x}\:\!(x-4)}

    Then I was going to rationalize the numerator by multiplying by 2+ \sqrt{x}
    but that seemed too complicated. . But did you try it?

    Where did i go wrong? . Nowhere . . . your work is correct!

    You would have: / \frac{2-\sqrt{x}}{2\sqrt{x}\;\!(x-4)}\cdot {\color{blue}\frac{2+\sqrt{x}}{2 + \sqrt{x}}} \;=\;\frac{4-x}{2\sqrt{x}\;\!(x-4)(2+\sqrt{x})} \;=\; \frac{-(x-4)}{2\sqrt{x}\;\!(x-4)(2+\sqrt{x})}

    . . . . . . . . . . . . . . =\;\frac{-({\color{red}\rlap{/////}}x-4)}{2\sqrt{x}\;\!({\color{red}\rlap{/////}}x-4)(2+\sqrt{x})} \;=\;\frac{-1}{2\sqrt{x}\;\!(2+\sqrt{x})}
    Thanks from Petrus
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