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This is the question:

$$\frac{1-2x-3x^2}{3x-x^2-5} \gt 0$$

What I did :

I got the answer as $$\left(x-3\right)\left(x+1\right) \gt 0$$

giving me the solution set : $x \in (-\infty,-1 )\cup(3,\infty)$
but the answer is $x \in (-\infty,-1 )\cup(\frac{1}{3},\infty)$

where am i going wrong?

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  • $\begingroup$ how did you get $(x-3)(x+1)$? $\endgroup$
    – danimal
    Commented Apr 21, 2015 at 14:17

2 Answers 2

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Just note $$\frac{1-2x-3x^2}{3x-x^2-5}>0\iff\frac{3x^2+2x-1}{x^2-3x+5}>0\iff\frac{(3x-1)(x+1)}{(x-\frac{3}{2})^2+\frac{11}{4}}>0$$ Since $(x-3/2)^2+11/4>0$ for every real number $x$ we need $(3x-1)(x+1)>0$, this is achieved iif $x\in(-\infty,-1)\cup(\frac{1}{3},\infty)$.

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  • $\begingroup$ Ya i did exactly the same thing. just factorised the numerator wrongly. THank you for your reply. iam such a fool. $\endgroup$
    – Max Payne
    Commented Apr 21, 2015 at 14:19
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The numerator is $-3x^2-2x+1=-(x+1)(3x-1)$ while the denominator is $-x^2+3x-5<0$.

Can you proceed from here?

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  • $\begingroup$ actually it was a factorising error of numerator (while applying quad formula) thanks :) $\endgroup$
    – Max Payne
    Commented Apr 21, 2015 at 14:21
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$
    – Mark Viola
    Commented Apr 21, 2015 at 15:05

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