Question

How to simplify this??

\[ \frac{-\sqrt{3} - 1}{1+(-\sqrt{3})*1}\]
\[ \frac{-\sqrt{3} - 1}{1-\sqrt{3}}\]
\[ \frac{-\sqrt{3} - 1}{1-\sqrt{3}} * \frac{1-\sqrt{3}} {1-\sqrt{3}} \]


\[ \frac{2}{2(2-\sqrt{3})} \]

Is this correct??

Answer 1

In step 3, you should multiply the expression by the denominator's conjugate, top and bottom.

Answer 2

That allows you to eliminate radical expressions in the denominator.

Answer 3

Could you show me? I don't see what you mean. I thought I did multiply the expression

Answer 4

\[\frac{ -\sqrt{3}-1 }{ 1-\sqrt{3} } \times \frac{ 1+\sqrt{3} }{ 1+\sqrt{3} } \]

Answer 5

That should be your step 3.

Answer 6

I did that and I get 2 for the numerator and \( 2(2-\sqrt{3})\) for the denominator

Answer 7

No, if you multiply a radical expression by its conjugate, the square roots disappear.

Answer 8

Yeah, I used the same I see that.

Answer 9

I use - and not the conjugate

Answer 10

\[(1-\sqrt{3})(1+\sqrt{3}) = 1-\sqrt{3} + \sqrt{3} - 3 = 2\]

Answer 11

Do you agree with my above expansion?

Answer 12

Yes

Answer 13

Ok, so 2 is now your denominator.

Answer 14

Let me do the top

Answer 15

Let's multiply the numerator now.

Answer 16

I got -2

Answer 17

One sec I made a boo boo

Answer 18

Check your numerator again. It should contain some square roots.

Answer 19

I have

\[ -\sqrt{3} -\sqrt{9} -1 -\sqrt{3} \] is this correct?

Answer 20

Yes, you can combine the terms to make it easier to read.

Answer 21

\[-2\sqrt{3}-4\]

Answer 22

Now that you have the numerator, don't forget that you still have a denominator, -2.

Answer 23

From what I put up where di the 4 come from?

Answer 24

Oh your \[- \sqrt{9} - 1 = -3 - 1 = - 4\]

Answer 25

Combine your numerator and your denominator, now what do you have?

Answer 26

Ok how about eh 2 and the sqr(3) ?? from what I put up.

Answer 27

Oh that's from \[-\sqrt{3}-\sqrt{3} = -2\sqrt{3}\]

Answer 28

\[\frac{ -2\sqrt{3}-4 }{ -2 } = \sqrt{3} + 2\]

Answer 29

Yep that is it. Thank you so much.

Answer 30

Are you a qualified helper?

Answer 31

I just like to help :)