# How to simplify $\frac{1}{10x}(16-2(-5x+8)^{\frac{1}{2}})$?

This is an easy question that I would like to get help with. How do you simplify this expression fully? $$\frac{1}{10x}(16-2(-5x+8)^{\frac{1}{2}})$$

Is it possible to go further?

EDIT:

The simplification is from the following question: Write the expression below so that the denominator does not contain any root expressions and simplify as far as possible:

$$-\frac{8}{5}<x<\frac{8}{5}, \quad \frac{\sqrt{8+5x}-\sqrt{8-5x}}{\sqrt{5x+8}+\sqrt{-5x+8}}=\frac{A}{B}$$

• Divide both numerator and denominator by 2. Feb 27, 2013 at 19:25

You can take the denominator into the square root, getting $$\frac{1}{10x}\left(16-2(-5x+8)^{\frac{1}{2}}\right)=\frac 8{5x}-\left(\frac 1{5x}+\frac 8{25x^2}\right)^\frac 12$$ Whether that is simpler is in the eye of the beholder.

Added: You have $$\frac{\sqrt{8+5x}-\sqrt{8-5x}}{\sqrt{5x+8}+\sqrt{-5x+8}}=\frac{(\sqrt{8+5x}-\sqrt{8-5x})(\sqrt{5x+8}-\sqrt{-5x+8})}{10x}=\\ \frac{16-2\sqrt {(8-5x)(8+5x)}}{10x}=\frac {16-2\sqrt{64-25x}}{10x}=\frac {8-\sqrt{64-25x}}{5x}$$

• Hi @Ross Millikan, please check my updated question and see if that changes things... Feb 27, 2013 at 19:36
• Hmm appearantly that is not correct. Any other ideas? Thank you for your help @Ross Millikan! Feb 27, 2013 at 19:58
• @LukasArvidsson: you dropped a factor $(8+5x)$ under the radical. Feb 27, 2013 at 20:50
• Thank you for pointing that out! Feb 27, 2013 at 21:58

From what I can see, no. The square root of the $x$ term completely ruins any chance of further simplification. You'll just end up with a rearrangement of how you've presented it.

1/10^X(16-2)(-5x+8)^1/2

1/10^X(14)(-5x+8)^1/2

1/10 (14)^x sqrt (-5x+8)

interesting problem

• Welcome to MSE! This is currently very difficult to understand. Can you add details to clarify? Also, it really helps to format using MathJax (see FAQ). Regards Aug 16, 2013 at 17:56