Thread: Simplifying with fractional exponets

1. Simplifying with fractional exponets

Simplify the expressions.

Problem 1:
$\displaystyle \frac{2(1+x)^{1/2} - x(1+x)^{-1/2}}{x+1}$

The final answer to problem 1 is:
$\displaystyle \frac{x+2}{(x+1)^{3/2}}$

Problem 2:
$\displaystyle \frac{(7-3x)^{1/2} + \frac{3}{2} x (7-3x)^{-1/2}}{7-3x}$

2. Re: Simplifying with fractional exponets

Hello, friesr!

$\displaystyle \text{Simplify:}$

. . $\displaystyle [1]\;\;\frac{2(x+1)^{\frac{1}{2}} - x(x+1)^{-\frac{1}{2}}}{x+1}$

Multiply by $\displaystyle \frac{(x+1)^{\frac{1}{2}}}{(x+1)^{\frac{1}{2}}}$

. . $\displaystyle \frac{(x+1)^{\frac{1}{2}}}{(x+1)^{\frac{1}{2}}} \cdot \frac{2(x+1)^{\frac{1}{2}} - x(x+1)^{-\frac{1}{2}}}{x+1}$

. . $\displaystyle =\;\frac{2(x+1)-x}{(x+1)^{\frac{3}{2}}} \;=\;\frac{x+2}{(x+1)^{\frac{3}{2}}}$

$\displaystyle [2]\;\frac{(7-3x)^{1/2} + \frac{3}{2} x (7-3x)^{-1/2}}{7-3x}$

Multiply by $\displaystyle \frac{2(7-3x)^{\frac{1}{2}}}{2(7-3x)^{\frac{1}{2}}}$

. . $\displaystyle \frac{2(7-3x)^{\frac{1}{2}}}{2(7-3x)^{\frac{1}{2}}}\cdot \frac{(7-3x)^{\frac{1}{2}} + \frac{3}{2}x(7-3x)^{-\frac{1}{2}}}{7-3x}$

. . $\displaystyle =\;\frac{2(7-3x) + 3x}{2(7-3x)^{\frac{3}{2}}} \;=\; \frac{14- 6x + 3x}{2(7-3x)^{\frac{3}{2}}} \;=\;\frac{14 - 3x}{2(7-3x)^{\frac{3}{2}}}$

3. Re: Simplifying with fractional exponets

Seeing how this works but not sure on what to choose to multiply thru by.

On this problem:

$\displaystyle \frac{2x(x+6)^{1/2} - x^{2}(4)(x+6)^{3}}{(x+6)^8}$

$\displaystyle \frac{(x+6)^{1/2}}{(x+6)^{1/2}} * \frac{2x(x+6)^{1/2} - x^{2}(4)(x+6)^{3}}{(x+6)^8}$

$\displaystyle \frac{2x(x+6) - x^{2}(4)(x+6)^{7/2}}{8(x+6)^{3/2}}$

$\displaystyle \frac{x(x+6) - x^{2}(2)(x+6)^{7/2}}{4(x+6)^{3/2}}$

It does not look like my choice is working out well.

4. Re: Simplifying with fractional exponets

You have:
$\displaystyle \frac{2x(x+6)^{\frac{1}{2}}-4x^2(x+6)^3}{(x+6)^8}$
$\displaystyle =\frac{(x+6)^{\frac{1}{2}}[2x-4x^2(x+6)^{\frac{5}{2}}]}{(x+6)^{\frac{1}{2}}(x+6)^{\frac{15}{2}}}$
$\displaystyle =\frac{2x-4x^2(x+6)^{\frac{5}{2}}}{(x+6)^{\frac{15}{2}}}$

5. Re: Simplifying with fractional exponets

$\displaystyle \frac{2x - 4x^{2}(x+6)^{5/2}}{(x+6)^{15/2}}$

We can factor out a 2x

$\displaystyle \frac{2x (1 - 2x(x+6)^{5/2}}{(x+6)^{15/2}}$

Can we factor out more (x+6)?

6. Re: Simplifying with fractional exponets

Originally Posted by friesr
$\displaystyle \frac{2x - 4x^{2}(x+6)^{5/2}}{(x+6)^{15/2}}$

We can factor out a 2x

$\displaystyle \frac{2x (1 - 2x(x+6)^{5/2}}{(x+6)^{15/2}}$

Can we factor out more (x+6)?
You can factor out (x+6) but I don't think it would be useful.

\displaystyle \begin{align*}\frac{2x - 4x^{2}(x+6)^{5/2}}{(x+6)^{15/2}} &=\frac{2x}{(x+6)^{15/2}} - \frac{4x^{2}(x+6)^{5/2}}{(x+6)^{15/2}} \\ &=\frac{2x}{(x+6)^{15/2}} - \frac{4x^{2}}{(x+6)^{5}}\end{align*}

This is as far as I can simplify.

Please see this link:simplify &#40;2x-4x&#94;2&#40;x&#43;6&... - Wolfram|Alpha