How do you rationalize the denominator and simplify $\frac{1}{1 - 8 \sqrt{2}}$ $1/(1-8sqrt2)$ ?

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Answer 1 · 2018-03-25T03:10:45

I believe this should be simplified as $\frac{- (8 \sqrt{2} + 1)}{127}$ $(-(8sqrt2+1))/127$ .

To rationalize the denominator, you must multiply the term that has the $\sqrt{}$ $sqrt$ by itself, to move it to the numerator. So:

$\Rightarrow$ $=>$ $\frac{1}{1 - 8 \cdot \sqrt{2}} \cdot 8 \sqrt{2}$ $1/(1-8*sqrt2)*8sqrt2$

This will give:

$\Rightarrow$ $=>$ $\frac{8 \sqrt{2} + 1}{1 - {(8 \sqrt{2})}^{2}}$ $(8sqrt2+1)/(1-(8sqrt2)^2$

${(8 \sqrt{2})}^{2} = 64 \cdot 2 = 128$ $(8sqrt2)^2=64*2=128$

$\Rightarrow$ $=>$ $\frac{8 \sqrt{2} + 1}{1 - 128}$ $(8sqrt2+1)/(1-128)$

$\Rightarrow$ $=>$ $\frac{8 \sqrt{2} + 1}{- 127}$ $(8sqrt2+1)/-127$

The negative cam also be moved to the top, for:

$\Rightarrow$ $=>$ $\frac{- (8 \sqrt{2} + 1)}{127}$ $(-(8sqrt2+1))/127$

Answer 2 · 2018-03-25T03:14:30

$\frac{- 1 - 8 \sqrt{2}}{127}$ $(-1-8sqrt2)/127$

Multiply the numerator and the denominator by the surd (to undo the surd) and the negative of the extra value.

$\frac{1}{1 - 8 \sqrt{2}}$ $1/(1-8sqrt2$ x $\frac{- 1 + 8 \sqrt{2}}{- 1 + 8 \sqrt{2}}$ $(-1+8sqrt2)/(-1+8sqrt2$

$\frac{1 (1 + 8 \sqrt{2})}{(1 - 8 \sqrt{2}) (1 + 8 \sqrt{2})}$ $(1(1+8sqrt2))/((1-8sqrt2)(1+8sqrt2)$

Expand brackets. Use the FOIL rule for the denominator.

$\frac{1 + 8 \sqrt{2}}{- 127}$ $(1+8sqrt2)/-127$

You could simplify further by taking the negative of the denominator and apply it to the numerator.

$\frac{- 1 - 8 \sqrt{2}}{127}$ $(-1-8sqrt2)/127$

How do you rationalize the denominator and simplify 11−8√21/(1-8sqrt2)?