anonymous
  • anonymous
Can I factor a quadratic w/ a neg. x^2?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Owlfred
  • 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!
amistre64
  • amistre64
you can, but its usually more effort than what its worth
myininaya
  • myininaya
it depends on the quad

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More answers

anonymous
  • anonymous
I'm trying to solve a limit question and it has two quadratics undera radical, so I have to simplify. \[2+ x - x^2\]
myininaya
  • myininaya
-x^2+x+2 -(x^2-x-2) -(x-2)(x+1)
anonymous
  • anonymous
thank you -- I still can't figure out my limit, but thanks
myininaya
  • myininaya
what is it?
anonymous
  • anonymous
Lim as X goes to -Sqrt (2\[\lim x \rightarrow -1 \sqrt{2+x-x ^{2 } /x ^{2+4x+3}?}\]
anonymous
  • 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!
myininaya
  • myininaya
wow that looks crazy
myininaya
  • myininaya
\[\lim_{x \rightarrow -1} \sqrt{\frac{2+x-x^2}{x^2+4x+3}}\]
myininaya
  • myininaya
is that right?
anonymous
  • anonymous
yes! Its confusing to me
myininaya
  • myininaya
\[\lim_{x \rightarrow -1}\sqrt{\frac{-1(x^2-x-2)}{(x+3)(x+1)}}=\lim_{x \rightarrow -1}\sqrt{\frac{-1(x+1)(x-2)}{(x+3)(x+1)}}\]
myininaya
  • myininaya
x+1 's cancel right! :)
myininaya
  • myininaya
\[\lim_{x \rightarrow -1}\sqrt{\frac{-1(x-2)}{x+3}}=\sqrt{\frac{-1(-1-2)}{-1+3}}=\sqrt{\frac{-1(-3)}{2}}=\sqrt{\frac{3}{2}}\]
myininaya
  • myininaya
\[\sqrt{\frac{3}{2}}=\frac{\sqrt{3}}{\sqrt{2}}*\frac{\sqrt2}{\sqrt2}=\frac{\sqrt{6}}{2}\]
anonymous
  • 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!!!!
myininaya
  • myininaya
lol you didn't see what i typed until you got to the thank part lol right?
anonymous
  • anonymous
Can I give you another tricky one?
myininaya
  • myininaya
yes of course

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