Question

These darn square roots.. Get me EVERY time.. help?

(picture)

Answer 1

this question right?

Answer 2

That was fast. & Yes :)

Answer 3

They want you to rationalize the denominator, do you know how to do this?

Answer 4

I do, and when I finished up I got (4sqrt2)15. but this is what the answer is supposed to look like.. (the picture)

Answer 5

Oops. That is supposed to say (4sqrt2)/15

Answer 6

Unfortunately \[\Large \frac{4\sqrt{2}}{15}\] is not correct

Answer 7

Yeah, once I wrote it down I noticed my mistake. But I tried it again and now it is an endless cycle of me trying to get rid of the squareroot. Can I get rid of them both at the same time?

Answer 8

Using the difference of squares rule, we know that


\[\Large (a+b)(a-b) = a^2 - b^2\]


If \(\Large a = \sqrt{7}\) and \(\Large b = \sqrt{8}\), then


\[\Large (a+b)(a-b) = a^2 - b^2\]

\[\Large (\sqrt{7}+\sqrt{8})(\sqrt{7}-\sqrt{8}) = \left(\sqrt{7}\right)^2 - \left(\sqrt{8}\right)^2\]

\[\Large (\sqrt{7}+\sqrt{8})(\sqrt{7}-\sqrt{8}) = 7 - 8\]

\[\Large (\sqrt{7}+\sqrt{8})(\sqrt{7}-\sqrt{8}) = -1\]


See how I'm using this trick to effectively get rid of the square roots?

Answer 9

So if you multiply both numerator and denominator by \(\Large \sqrt{7}-\sqrt{8} \), then you'll rationalize the denominator

Answer 10

Oh okay, so what you get for the numerator, you just then make negative because you are dividing by -1?

Answer 11

exactly, so what do you get for the final answer?

Answer 12

I got 10sqrt2 - 5sqrt7

Answer 13

for your final answer or just for the numerator?

Answer 14

before you divide by -1

Answer 15

just the numerator?

Answer 16

basically, what do you get when you multiply \(\Large \sqrt{7}-5 \) by \(\Large \sqrt{7}-\sqrt{8} \)

Answer 17

I got 10sqrt2 - 5sqrt7

Answer 18

ah ok, something is a bit off, one sec

Answer 19

Wait, I meant 4sqrt7 if that changes anything..

Answer 20

\[\Large \left(\sqrt{7}-5\right)\left(\sqrt{7}-\sqrt{8}\right)\]

\[\Large \sqrt{7}\left(\sqrt{7}-\sqrt{8}\right)-5\left(\sqrt{7}-\sqrt{8}\right)\]

\[\Large \sqrt{7}*\sqrt{7}-\sqrt{7}*\sqrt{8}-5\sqrt{7}+5\sqrt{8}\]

\[\Large \sqrt{7*7}-\sqrt{7*8}-5\sqrt{7}+5\sqrt{8}\]

\[\Large \sqrt{49}-\sqrt{56}-5\sqrt{7}+5\sqrt{8}\]

\[\Large 7-\sqrt{56}-5\sqrt{7}+5\sqrt{8}\]

\[\Large 7-\sqrt{4*14}-5\sqrt{7}+5\sqrt{8}\]

\[\Large 7-\sqrt{4}*\sqrt{14}-5\sqrt{7}+5\sqrt{8}\]

\[\Large 7-2\sqrt{14}-5\sqrt{7}+5\sqrt{8}\]


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So

\[\Large \left(\sqrt{7}-5\right)\left(\sqrt{7}-\sqrt{8}\right)=7-2\sqrt{14}-5\sqrt{7}+5\sqrt{8}\]

Answer 21

This means

\[\Large \frac{\sqrt{7}-5}{\sqrt{7}+\sqrt{8}}\]

becomes

\[\Large \frac{7-2\sqrt{14}-5\sqrt{7}+5\sqrt{8}}{-1}\]

\[\Large -7+2\sqrt{14}+5\sqrt{7}-5\sqrt{8}\]

Answer 22

Oh I just multiplied it with the 5. I didn't distribute it correctly.. Whoops

Answer 23

that's ok, happens to everyone

Answer 24

Well thank you for clearing that up :) I really appreciate it !!!

Answer 25

you're welcome