# nth roots: single quotient in which positive exponents appear

• April 17th 2010, 05:39 PM
desiderius1
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!
• April 17th 2010, 06:07 PM
Gusbob
Quote:

Originally Posted by desiderius1
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?
• April 17th 2010, 06:14 PM
desiderius1
Write the expression as a single quotient in which positive exponents and/or radicals appear. Cannot rationalize the denominator.
• April 17th 2010, 06:28 PM
Gusbob
Quote:

Originally Posted by desiderius1
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.
• April 17th 2010, 07:00 PM
desiderius1
$
\frac{\sqrt{5x+7}}{2\sqrt{x-6}} + \frac{\sqrt{x-6}}{3\sqrt{5x+7}}
$

Numerator:
$
\sqrt{5x+7} * 3\sqrt{5x+7} = 3(5x+7) = 15x + 21
$

$
\sqrt{x-6} * 2\sqrt{x-6} = 2 (x-6) = 2x - 12
$

$
15x + 21 + 2x - 12 = 17x + 9
$

Denominator:
$
2\sqrt{x-6} * 3\sqrt{5x+7} = 6\sqrt{x-6}\sqrt{5x+7}
$

$
\frac{17x + 9}{6\sqrt{x-6}\sqrt{5x+7}}
$
• April 17th 2010, 07:55 PM
desiderius1
Another problem I'm having trouble with:

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

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

so far I'm here..
$
\frac{\sqrt{x+6} - \frac{3x}{\sqrt{x+6}}}{x+6}
$
• April 17th 2010, 08:31 PM
Gusbob
Quote:

Originally Posted by desiderius1
Another problem I'm having trouble with:

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

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

so far I'm here..
$
\frac{\sqrt{x+6} - \frac{3x}{\sqrt{x+6}}}{x+6}
$

Let the 2 terms on top have a common denominator:

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

Then you have:

$\frac{\sqrt{x+6} - \frac{3x}{\sqrt{x+6}}}{x+6}
=\frac{x+6 - 3x}{\sqrt{x+6}} \times \frac{1}{x+6}$

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

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

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

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

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..
• April 18th 2010, 12:07 AM
earboth
Quote:

Originally Posted by desiderius1
And another one..

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

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

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

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)
• April 18th 2010, 12:28 AM
desiderius1
$\frac{x+4}{1} + \frac{3}{2}x = \frac{2x+8}{2}+\frac{3x}{2}$

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

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

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

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

Cool thanks for helping.