What am I doing wrong dividing $\frac{3-\sqrt{-8}}{1+\sqrt{-2}}$?

Question

I took the negatives out of the square roots by making the equation: $$\frac{3-i\sqrt{8}}{1+i\sqrt{2}}$$

Then I multiplied it by the conjugate of the denominator: $$\frac{3-i\sqrt{8}}{1+i\sqrt{2}}*\frac{1-i\sqrt{2}}{1-i\sqrt{2}}$$

And then I got: $$\frac{3-3i\sqrt{2}-i\sqrt{8}+\sqrt{16}}{3}$$

Then, I knew that $\sqrt{8}$ is $2\sqrt{2}$, and then I knew that I could combine the $-2i\sqrt{2}$ with the $-3i\sqrt{2}$ and ended up with: $$\frac{3-6i\sqrt{2}\pm4}{3}$$

I know this isn't correct, so where did I go wrong?

Edit: Okay, so far, it looks like the the final answer is $$\frac{3-5i\sqrt{2}-4}{3}$$

I thought that anytime you take a square root, it has to be plus or minus. Is that not correct? — Garrett Smith, Aug 22 '16 at 23:56
$\sqrt{16}$ is unambiguous, meaning the positive square root. $x^2 = 16$ is different. — , Aug 22 '16 at 23:56
@GarrettSmith: No, that's not correct. We agree by definition that $\sqrt{16}=4.$ What you're confused by is the fact that there are two solutions to $x^{2}=16,$ namely $x=4$ and $x=-4.$ See this question. — Will R, Aug 22 '16 at 23:57
Right, the problem is that when I check my answer on Mathway.com, it says that the answer should be $-3+3\sqrt{2}+2i\sqrt{2}-4i$. According to this, even if I did say that it was only +4, I don't think I would get the correct answer. — Garrett Smith, Aug 22 '16 at 23:59
It should be minus (not plus) $\sqrt{16}$. And sqrt(16) is 4 because we were assuming $\sqrt 2$ and $\sqrt 8$ were positive. — fleablood, Aug 23 '16 at 00:05
Alright, well I wonder if I should report a bug to Mathway. Anyway, thanks for the help. — Garrett Smith, Aug 23 '16 at 00:08

score 2 · Accepted Answer · answered Aug 23 '16 at 00:02

2

Should be $-\sqrt 16$, also $-3-2=-5,$ not $-6$. Square roots that are included in the problem statement usually assume the positive root.

answered Aug 23 '16 at 00:02

rikhavshah

1,355

score 1 · Answer 2 · answered Aug 23 '16 at 00:03

1

You have done the multiplication wrong. $$(3-i\sqrt8)(1-i\sqrt2) = 3 - 2i\sqrt2 - 3i\sqrt2 \color{red}{-}\sqrt{16} = -1 -5i\sqrt 2$$

answered Aug 23 '16 at 00:03

Teoc

8,700

What am I doing wrong dividing $\frac{3-\sqrt{-8}}{1+\sqrt{-2}}$?

2 Answers2