Question

2a^2-4a+2/3a^2-3
Please show all work, and state any excuded values.

Answer 1

first, use quadratic formula to solve numerator
\[\huge 2a^2-4a+2\]
\[\huge 2a^2-2a-2a+2\]
\[\huge 2a(a-1)-2(a-1)\]
\[\huge (a-1)(2a-2)\]

Answer 2

\[(2a ^{2}-4a+2)/3a ^{2}-3\]=\[2(a ^{2}-2a+1)/3(a ^{2}-1)=2(a-1)^{2}/3(a-1)(a+1)=2(a-1)/3(a+1)\]
now if this is equal to 0..then a=1...and anot equal to -1 as the denominator will then become infinite

Answer 3

then, solve denominator,
\[\huge 3a^2-3\]
\[\huge 3(a^2-1)\]
\[\huge 3(a+1)(a-1)\]
Then, fraction becomes,
\[\huge \frac{(a-1)(2a-2)}{3(a+1)(a-1)}\]
\[\huge \frac{2a-2}{3(a+1)}=\frac{2a-2}{3a+3}\]

Answer 4

got it @Pyromancer

Answer 5

@RaGhavv your answer and my answer is same , but only difference is that when simplify your answer then your answer=my answer.
And i think that this is the final answer.

Answer 6

yeah! but he asked for excluded values so i thought it might be an equation...if not ,then also a is not equal to -1 i just added that...and yes both of our answer are indeed correct my friend..:)

Answer 7

Isn't there a excluded value?

Answer 8

@RaGhavv @mayankdevnani

Answer 9

yes there is...a is not equal to that value which makes the denominator infinite..

Answer 10

ok, and for the final answer i got 2(a-1)/3(a+1)

Answer 11

you are correct!! @Pyromancer

Answer 12

yes and so \[3(a+1)\neq0..\]

Answer 13

i got:
\[a\]

Answer 14

\[a \neq-1\]

Answer 15

@raghavv

Answer 16

@mayankdevnani

Answer 17

yes a is not equal to -1 and rest it can take anyother value..