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??

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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??

Mathematics
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In step 3, you should multiply the expression by the denominator's conjugate, top and bottom.
That allows you to eliminate radical expressions in the denominator.
Could you show me? I don't see what you mean. I thought I did multiply the expression

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\[\frac{ -\sqrt{3}-1 }{ 1-\sqrt{3} } \times \frac{ 1+\sqrt{3} }{ 1+\sqrt{3} } \]
That should be your step 3.
I did that and I get 2 for the numerator and \( 2(2-\sqrt{3})\) for the denominator
No, if you multiply a radical expression by its conjugate, the square roots disappear.
Yeah, I used the same I see that.
I use - and not the conjugate
\[(1-\sqrt{3})(1+\sqrt{3}) = 1-\sqrt{3} + \sqrt{3} - 3 = 2\]
Do you agree with my above expansion?
Yes
Ok, so 2 is now your denominator.
Let me do the top
Let's multiply the numerator now.
I got -2
One sec I made a boo boo
Check your numerator again. It should contain some square roots.
I have \[ -\sqrt{3} -\sqrt{9} -1 -\sqrt{3} \] is this correct?
Yes, you can combine the terms to make it easier to read.
\[-2\sqrt{3}-4\]
Now that you have the numerator, don't forget that you still have a denominator, -2.
From what I put up where di the 4 come from?
Oh your \[- \sqrt{9} - 1 = -3 - 1 = - 4\]
Combine your numerator and your denominator, now what do you have?
Ok how about eh 2 and the sqr(3) ?? from what I put up.
Oh that's from \[-\sqrt{3}-\sqrt{3} = -2\sqrt{3}\]
\[\frac{ -2\sqrt{3}-4 }{ -2 } = \sqrt{3} + 2\]
Yep that is it. Thank you so much.
Are you a qualified helper?
I just like to help :)

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