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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:

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

Thank you for your help!

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    $\begingroup$ Divide both numerator and denominator by 2. $\endgroup$ Commented Feb 27, 2013 at 19:25

3 Answers 3

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

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

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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

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    $\begingroup$ 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 $\endgroup$
    – Amzoti
    Commented Aug 16, 2013 at 17:56

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