Question

help?

Answer 1

@e.mccormick @Abhisar @Help12356789

Answer 2

Well, do you know what is needed to add or subtract radicals?

Answer 3

no :

Answer 4

It is like fractions. What is under needs to match.

\(\dfrac{2}{x}\pm \dfrac{1}{y}\) can't be added or subtracted, but \(\dfrac{2}{x}\pm \dfrac{1}{x}\) can. Well, \(2\sqrt{x}\pm \sqrt{y}\) can't be added or subtracted, but \(2\sqrt{x}\pm \sqrt{x}\) can. The key is that what is under the fraction or under the radical matches.

Your don't have matching things, but can one be changed so that they do match?

Answer 5

okay so it would be B?

Answer 6

It is like fractions. What is under needs to match.

\(\dfrac{2}{x}\pm \dfrac{1}{y}\) can't be added or subtracted, but \(\dfrac{2}{x}\pm \dfrac{1}{x}\) can. Well, \(2\sqrt{x}\pm \sqrt{y}\) can't be added or subtracted, but \(2\sqrt{x}\pm \sqrt{x}\) can.



The way to get things out from under a root is with prime factors. Take \(\sqrt{45}\) as an example. What do you get if you prime factor it? Well, 9*5=45, so \(\sqrt{9\cdot 5}\). But that is not prime factors. 3*3=9, so \(\sqrt{3\cdot 3\cdot 5}\) is using primes.

Now, because this is a square root, you have to pull things out in pairs. If it was a cube root it would be in triples and so on. Now, there are a pair of threes.

\(\sqrt{3\cdot 3\cdot 5}= 3 \sqrt{5}\)



Hehe, and here I was working that up. What one was b? I did not leave it open.

Answer 7

It is rational and equal to 0.

Answer 8

Yah, that is the correct one.