## anonymous 5 years ago Can I factor a quadratic w/ a neg. x^2?

1. Owlfred

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!

2. amistre64

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

3. myininaya

4. anonymous

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

5. myininaya

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

6. anonymous

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

7. myininaya

what is it?

8. anonymous

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

9. anonymous

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

10. myininaya

wow that looks crazy

11. myininaya

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

12. myininaya

is that right?

13. anonymous

yes! Its confusing to me

14. myininaya

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

15. myininaya

x+1 's cancel right! :)

16. myininaya

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

17. myininaya

$\sqrt{\frac{3}{2}}=\frac{\sqrt{3}}{\sqrt{2}}*\frac{\sqrt2}{\sqrt2}=\frac{\sqrt{6}}{2}$

18. anonymous

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!!!!

19. myininaya

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

20. anonymous

Can I give you another tricky one?

21. myininaya

yes of course