Question

More help on algebra please?

Answer 1

Rationalize the denominator and simplify \[\frac{ \sqrt{a+1}-2 }{ \sqrt{a+1}+2 }\] Once again I know I need to multiply it by the conjugate but again I need help doing the actual math.
I multiplied it by \[\frac{ \sqrt{a+1}-2 }{ \sqrt{a+1}-2 }\] and I got \[\frac{ 1-\sqrt{a+1} }{ -1 }\] but it doesn't fit where I need to put the answer.

Answer 2

sigh it's nice to know that someone is doing latex.. it sucks that my fraction latex is broken .

Answer 3

latex?

Answer 4

\[\frac{ \sqrt{a+1}-2 }{ \sqrt{a+1}+2 } \cdot \frac{ \sqrt{a+1}-2 }{ \sqrt{a+1}-2 } \]

Answer 5

yeah that's used to type Math. It's called Latex. But on my end it's coo coo

Answer 6

Ohh

Answer 7

\[\frac{ \sqrt{a+1}-2 }{ \sqrt{a+1}+2 } \cdot \frac{ \sqrt{a+1}-2 }{ \sqrt{a+1}-2 } \rightarrow \frac{(\sqrt{a+1}-2)(\sqrt{a+1}-2)}{(\sqrt{a+1}+2)(\sqrt{a+1}-2)}\]

Answer 8

can we not give direct answers please? It's against code of conduct

Answer 9

oh sorry I'm new here

Answer 10

deleted

Answer 11

ah. I see. Ok.

Answer 12

@LilyMQ I do need to know how

Answer 13

You can read more about the code of conduct here. http://openstudy.com/code-of-conduct

Answer 14

anyway where was I?

\[\frac{ \sqrt{a+1}-2 }{ \sqrt{a+1}+2 } \cdot \frac{ \sqrt{a+1}-2 }{ \sqrt{a+1}-2 } \rightarrow \frac{(\sqrt{a+1}-2)(\sqrt{a+1}-2)}{(\sqrt{a+1}+2)(\sqrt{a+1}-2)}\]

Answer 15

so we need to use the foil method... as usual the denominator is always easier because the middle terms are cut out

Answer 16

\[\frac{(\sqrt{a+1}-2)(\sqrt{a+1}-2)}{(\sqrt{a+1}+2)(\sqrt{a+1}-2)} =\frac{?}{(a+1)-4}\]

Answer 17

so foil is needed on the numerator .

Answer 18

our denominator should be a+1-4=a-3

Answer 19

dear gawd latex... U SUCK!