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Thread: nth roots: single quotient in which positive exponents appear

  1. April 17th 2010, 05:39 PM #1
    desiderius1
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    nth roots: single quotient in which positive exponents appear

    Hello, I'm stumped on this question and need some help.

    \sqrt{5x+7}\frac{1}{2\sqrt{x-6}} + \sqrt{x-6}\frac{1}{3\sqrt{5x+7}} \qquad x > 6

    so far i've done:
    \frac{\sqrt{5x+7}}{2\sqrt{x-6}} + \frac{\sqrt{x-6}}{3\sqrt{5x+7}}?

    Is this correct and if so, what do I do next?

    Thanks!
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  2. April 17th 2010, 06:07 PM #2
    Gusbob
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    Quote Originally Posted by desiderius1 View Post
    Hello, I'm stumped on this question and need some help.

    \sqrt{5x+7}\frac{1}{2\sqrt{x-6}} + \sqrt{x-6}\frac{1}{3\sqrt{5x+7}} \qquad x > 6

    so far i've done:
    \frac{\sqrt{5x+7}}{2\sqrt{x-6}} + \frac{\sqrt{x-6}}{3\sqrt{5x+7}}?

    Is this correct and if so, what do I do next?

    Thanks!
    What are you trying to do with this expression? Simplify it? Equate to 0 and solve for x?
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  3. April 17th 2010, 06:14 PM #3
    desiderius1
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    Write the expression as a single quotient in which positive exponents and/or radicals appear. Cannot rationalize the denominator.
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  4. April 17th 2010, 06:28 PM #4
    Gusbob
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    Quote Originally Posted by desiderius1 View Post
    Write the expression as a single quotient in which positive exponents and/or radicals appear. Cannot rationalize the denominator.
    Try multiplying both parts by the conjugate/conjugate first then cross multiply to make a common denominator.
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  5. April 17th 2010, 07:00 PM #5
    desiderius1
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    <br /> \frac{\sqrt{5x+7}}{2\sqrt{x-6}} + \frac{\sqrt{x-6}}{3\sqrt{5x+7}}<br />

    Numerator:
    <br /> \sqrt{5x+7} * 3\sqrt{5x+7} = 3(5x+7) = 15x + 21<br />

    <br /> \sqrt{x-6} * 2\sqrt{x-6} = 2 (x-6) = 2x - 12<br />

    <br /> 15x + 21 + 2x - 12 = 17x + 9<br />

    Denominator:
    <br /> 2\sqrt{x-6} * 3\sqrt{5x+7} = 6\sqrt{x-6}\sqrt{5x+7}<br />

    Answer:
    <br /> \frac{17x + 9}{6\sqrt{x-6}\sqrt{5x+7}}<br />
    Last edited by desiderius1; April 17th 2010 at 07:21 PM. Reason: Figured it out, had it wrong.
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  6. April 17th 2010, 07:55 PM #6
    desiderius1
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    Another problem I'm having trouble with:

    <br /> \frac{x+6^{1/2} - 3x(x+6)^{-1/2}}{x+6}, \quad x > -6<br />

    From the back of the book, the answer is..
    <br /> \frac{6-2x}{(x+6)^{3/2}}<br />

    so far I'm here..
    <br /> \frac{\sqrt{x+6} - \frac{3x}{\sqrt{x+6}}}{x+6}<br />
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  7. April 17th 2010, 08:31 PM #7
    Gusbob
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    Quote Originally Posted by desiderius1 View Post
    Another problem I'm having trouble with:

    <br /> \frac{x+6^{1/2} - 3x(x+6)^{-1/2}}{x+6}, \quad x > -6<br />

    From the back of the book, the answer is..
    <br /> \frac{6-2x}{(x+6)^{3/2}}<br />

    so far I'm here..
    <br /> \frac{\sqrt{x+6} - \frac{3x}{\sqrt{x+6}}}{x+6}<br />
    Let the 2 terms on top have a common denominator:

    Numerator: \frac{x+6 - 3x}{\sqrt{x+6} }

    Then you have:

    \frac{\sqrt{x+6} - \frac{3x}{\sqrt{x+6}}}{x+6}<br /> =\frac{x+6 - 3x}{\sqrt{x+6}} \times \frac{1}{x+6}

    = \frac{6-2x}{(x+6)^{3/2}}
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  8. April 17th 2010, 11:56 PM #8
    desiderius1
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    And another one..

    Factor the given expression. Express your answer so that only positive exponents occur.

    <br /> (x+4)^{3/2} + x \times \frac{3}{2}(x+4)^{1/2} \qquad x >= -4<br />

    The answer is:
    <br /> \frac{1}{2}(x+4)^{1/2}(5x+8)<br />

    Here is where I am:
    Factor 1/2 from the exponents..
    \frac{1}{2}(x+4)^{2/2}+\frac{3}{2}x(x+4)

    Multiply the 3/2x(x+4)
    \frac{1}{2}(x+4)^{2/2}+\frac{3}{2}x^2+6x)

    I don't think that is right..
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  9. April 18th 2010, 12:07 AM #9
    earboth
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    Quote Originally Posted by desiderius1 View Post
    And another one..

    Factor the given expression. Express your answer so that only positive exponents occur.

    <br /> (x+4)^{3/2} + x \times \frac{3}{2}(x+4)^{1/2} \qquad x >= -4<br />

    The answer is:
    <br /> \frac{1}{2}(x+4)^{1/2}(5x+8)<br />

    Here is where I am:
    Factor 1/2 from the exponents..
    \frac{1}{2}(x+4)^{2/2}+\frac{3}{2}x(x+4)

    Multiply the 3/2x(x+4)
    \frac{1}{2}(x+4)^{2/2}+\frac{3}{2}x^2+6x)

    I don't think that is right..
    1. Split (x+4)^{\frac32} = (x+4) \cdot (x+4)^{\frac12}

    2. Now you can factor out (x+4)^{\frac12} :

    (x+4)^{\frac12} \cdot \left((x+4) + \frac32 x \right)

    3. Simplify! (The final result is 2 steps ahead)
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  10. April 18th 2010, 12:28 AM #10
    desiderius1
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    \frac{x+4}{1} + \frac{3}{2}x = \frac{2x+8}{2}+\frac{3x}{2}

    = \frac{5x+8}{2}

    = (x+4)^{1/2} \cdot \frac{5x+8}{2}

    = (x+4)^{1/2}(5x+8) \cdot \frac{1}{2}

    = \frac{1}{2}(x+4)^{1/2}(5x+8)

    Cool thanks for helping.
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