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I'm asked to simplify $\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2}$ and am provided with the solution $\frac{-5\sqrt{2}-6}{7}$

I have tried several approaches and failed. Here's one path I took:

(Will try to simplify the left hand side fraction part first and then deal with the $-2^{1/2}$ later)

$\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}$ The root of 16 is 4 and the root of 8 could be written as $2\sqrt{2}$ thus:

$\frac{2\sqrt{2}-4}{4-\sqrt{2}}$

Not really sure where to go from here so I tried multiplying out the radical in the denominator:

$\frac{2\sqrt{2}-4}{4-\sqrt{2}}$ = $\frac{2\sqrt{2}-4}{4-\sqrt{2}} * \frac{4+\sqrt{2}}{4+\sqrt{2}}$ = $\frac{(2\sqrt{2}-4)(4+\sqrt{2})}{16-2}$ =

(I become less certain in my working here)

$\frac{8\sqrt{2}*2(\sqrt{2}^2)-16-4\sqrt{2}}{14}$ = $\frac{8\sqrt{2}*4-16-4\sqrt{2}}{14}$ = $\frac{32\sqrt{2}-16-4\sqrt{2}}{14}$ = $\frac{28\sqrt{2}-16}{14}$

Then add back the $-2^{1/2}$ which can also be written as $\sqrt{2}$

This is as far as I can get. I don't know if $\frac{28\sqrt{2}-16}{14}-\sqrt{2}$ is still correct or close to the solution. How can I arrive at $\frac{-5\sqrt{2}-6}{7}$?

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    $\begingroup$ Did you mean $8\sqrt{2} \ast 2 \ast (\sqrt{2})^2$? I think you should have two terms here $\endgroup$ Jan 7 '19 at 2:48
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You were doing fine until the place where you tried to expand $(2\sqrt2 - 4)(4 + \sqrt2).$

There are mnemonic techniques for this but I think plain old distributive law works well enough: \begin{align} (2\sqrt2 - 4)(4 + \sqrt2) &= (2\sqrt2 - 4)4 + (2\sqrt2 - 4)\sqrt2 \\ &= (8\sqrt2 - 16) + (4 - 4\sqrt2) \\ &= 4\sqrt2 - 12. \end{align}

Next you might notice a chance to cancel a factor of $2$ in the numerator and denominator of $\frac{4\sqrt2 - 12}{14}.$

And finally you'll want to change the $-\sqrt2$ so that you have two fractions with a common denominator and can finish.

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  • $\begingroup$ Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here? $\endgroup$
    – Doug Fir
    Jan 7 '19 at 3:38
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    $\begingroup$ It's the same rule you would apply to simplify something like $\frac37 - 2.$ The $2$ is equal to $\frac21,$ which is equal to $\frac{7\cdot2}{7\cdot1}.$ In your problem you have $\sqrt2$ instead of $2$ but the principle is the same. $\endgroup$
    – David K
    Jan 7 '19 at 3:42
  • $\begingroup$ Hi David, thanks for clarifying that, I understand it now $\endgroup$
    – Doug Fir
    Jan 7 '19 at 4:01
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$$\begin{align} \frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}-\sqrt{2}&=\frac{2\sqrt{2}-4}{4-\sqrt{2}}-\sqrt{2}\\ &=\frac{2\sqrt{2}-4}{4-\sqrt{2}}-\frac{4\sqrt{2}-2}{4-\sqrt{2}}\\ &=\frac{-2\sqrt{2}-2}{4-\sqrt{2}}\\ &=\frac{-2\sqrt{2}-2}{4-\sqrt{2}}~\cdot~\frac{4+\sqrt{2}}{4+\sqrt{2}}\\ &=\frac{-10\sqrt{2}-12}{14}\\ &=\frac{-5\sqrt{2}-6}{7} \end{align}$$

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  • $\begingroup$ Thank you for the answer. I'm trying to understand what you are doing on the first new line, where you subtract the fraction $\frac{4\sqrt{2}-2}{4\sqrt{2}-2}$. 1. What's the objective here and 2. Why is the new denominator unchanged, since the next lines denominator is the same, $4-\sqrt{2}$? $\endgroup$
    – Doug Fir
    Jan 7 '19 at 3:12
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\begin{align} \frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2} & = \frac{2\sqrt{2}-4}{4-\sqrt{2}}\cdot \frac{4+\sqrt{2}}{4+\sqrt{2}} - \sqrt{2} \\ & = \frac{4\sqrt{2}-12}{14} - \sqrt{2} \\ & = \frac{2\sqrt{2}-6}{7} - \sqrt{2} \\ & = \frac{2\sqrt{2}-6 -7 \sqrt{2}}{7}\\ & = \frac{-5\sqrt{2} -6 }{7} \end{align}

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