Question

Can I factor a quadratic w/ a neg. x^2?

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

you can, but its usually more effort than what its worth

Answer 3

it depends on the quad

Answer 4

I'm trying to solve a limit question and it has two quadratics undera radical, so I have to simplify. \[2+ x - x^2\]

Answer 5

-x^2+x+2
-(x^2-x-2)
-(x-2)(x+1)

Answer 6

thank you -- I still can't figure out my limit, but thanks

Answer 7

what is it?

Answer 8

Lim as X goes to -Sqrt (2\[\lim x \rightarrow -1 \sqrt{2+x-x ^{2 } /x ^{2+4x+3}?}\]

Answer 9

the denominator under the radical is x^2 +4x +3
I couldn't get it to read the right way in the editor. Thanks!

Answer 10

wow that looks crazy

Answer 11

\[\lim_{x \rightarrow -1} \sqrt{\frac{2+x-x^2}{x^2+4x+3}}\]

Answer 12

is that right?

Answer 13

yes! Its confusing to me

Answer 14

\[\lim_{x \rightarrow -1}\sqrt{\frac{-1(x^2-x-2)}{(x+3)(x+1)}}=\lim_{x \rightarrow -1}\sqrt{\frac{-1(x+1)(x-2)}{(x+3)(x+1)}}\]

Answer 15

x+1 's cancel right! :)

Answer 16

\[\lim_{x \rightarrow -1}\sqrt{\frac{-1(x-2)}{x+3}}=\sqrt{\frac{-1(-1-2)}{-1+3}}=\sqrt{\frac{-1(-3)}{2}}=\sqrt{\frac{3}{2}}\]

Answer 17

\[\sqrt{\frac{3}{2}}=\frac{\sqrt{3}}{\sqrt{2}}*\frac{\sqrt2}{\sqrt2}=\frac{\sqrt{6}}{2}\]

Answer 18

yes -- got that
great! Thats it! They have the answer as being sqare root of 6 /2. How do they get it in that form?
Ah -- Thank you!!!!

Answer 19

lol you didn't see what i typed until you got to the thank part lol
right?

Answer 20

Can I give you another tricky one?

Answer 21

yes of course