Question

I need help on solving inequalities of the form
[f(x)/g(x)] > 0 where f(x) and g(x) are quadratic expressions that are either factorisable or always positive. http://dl.dropbox.com/u/63664351/Function%20and%20graphs%201.PNG

Answer 1

does it really say "find the range"?

Answer 2

I think you just solve it like normal inequality question.

Answer 3

yeah i think some algebra needed on the left. rationalize denominator perhaps

Answer 4

algebra a little slow this morning

Answer 5

ok apparently the left hand side is \(2x+2\sqrt{2x+1}+2\) so now it is easy enough
\[2x+2\sqrt{2x+1}+2<2x+9\]
\[\sqrt{2x+1}<\frac{9}{2}\]
\[2x+1<\frac{49}{4}\]
\[x<\frac{45}{8}\] and of course the domain requires \(x\geq-\frac{1}{2}\)

Answer 6

typo there, should be
\[\sqrt{2x+1}\leq \frac{7}{2}\]

Answer 7

Hey thanks but I think you need x>0 as well. Otherwise the fraction become error.

Answer 8

why \(x>0\)?

Answer 9

ooooooooooooh i see
actually it is not \(x>0\) but rather \(x\neq 0\)

Answer 10

No if you make x<0 then the root become imaginary

Answer 11

Wait. Let me think

Answer 12

so \((-\frac{1}{2},0)\cup (0,\frac{45}{8})\)

Answer 13

no you are still good between \( -\frac{1}{2}\) and 0

Answer 14

Lets step a back for a bit. How did you re-organize the left hand side. Nice work though.

Answer 15

actually to be more precise, it is
\[[-\frac{1}{2},0)\cup (0,\frac{45}{8})\]

Answer 16

multiply out in the denominator, combine like terms, rationalize denominator

Answer 17

gotta run, back later

Answer 18

How do you multiply out the denominator like this one?

Answer 19

what do you multiply it with?

Answer 20

\[(1-\sqrt{2x+1})^2=(1-\sqrt{2x+1})(1-\sqrt{2x+1})=1-2\sqrt{2x+1}+2x+1\]
\[=2x+2-2\sqrt{2x+1}\] giving you
\[\frac{4x^2}{2x+2-2\sqrt{2x+1}}=\frac{2x^2}{x+1-\sqrt{2x+1}}\]

Answer 21

then to rationalize, multiply top and bottom by the conjugate, namely
\[x+1+\sqrt{2x+1}\]

Answer 22

numerator is \(2x^2(x+1+\sqrt{2x+1})\) and a miracle occurs in the denominator, you get \(x^2\) cancelling gives the result

Answer 23

Wow. Thanks so much! I wonder if you have an application/ web application to instantly calculate something like this. I try wolfram alpha but it gave me terrible result http://www.wolframalpha.com/input/?i=4x%5E2%2F%281-sqrt%281-2x%29%29%5E2%3C2x%2B9

Answer 24

this works, but i think your problem has \(\sqrt{2x+1}\) right?

Answer 25

at least i hope so, otherwise i solved the wrong problem

Answer 26

ah yeah! thx :DD

Answer 27

http://www.wolframalpha.com/input/?i=4x^2%2F%281-sqrt%281%2B2x%29%29^2%3C2x%2B9

Answer 28

look it is right

Answer 29

Hey hang on! why the domain require x=-1/2?
After all if x=-1/2 , the denominator will become (1-1)^2 = 0.
You can't have denominator zero!

Answer 30

LOL nevermind.