Trying to simplify $\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2}$ into $\frac{-5\sqrt{2}-6}{7}$

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 $$\frac{-5\sqrt{2}-6}{7}$$?

• Did you mean $8\sqrt{2} \ast 2 \ast (\sqrt{2})^2$? I think you should have two terms here Jan 7 '19 at 2:48

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.

• Hi David. Regarding your last sentence, would it be possible to spell that out for me? What's the rule here? Jan 7 '19 at 3:38
• 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. Jan 7 '19 at 3:42
• Hi David, thanks for clarifying that, I understand it now Jan 7 '19 at 4:01

\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}

• 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, $4-\sqrt{2}$? Jan 7 '19 at 3:12

\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}