# Inequalities- tricky question

1. Aug 15, 2011

### J-Girl

Inequalities- tricky question!!

hiii,
i was wondering if anybody knew how to help me with this one tricky homework question. i can do most of the inequaliies ive come across, but how do you solve an inequality if you cant factor it???
the question is: (x^2-4x+7)/(x^2+x-6)
i know that the denominator is easily factorable ((x+3)(x-2)) but the numerator isnt? help plzzzz ive been stuck on this for ages!!!:(:(

2. Aug 15, 2011

### disregardthat

Re: Inequalities- tricky question!!

Rewrite (x^2-4x+7) to (x-2)^2+3. What can you say about this?

3. Aug 15, 2011

### Mentallic

Re: Inequalities- tricky question!!

Ok so I'm guessing by inequalities you mean something like $$\frac{x^2-4x+7}{x^2+x-6}>1$$ for example?

Try multiplying both sides by $$(x^2+x-6)^2$$ since you know this has to be a non-negative number, and don't expand! Use your knowledge of factorizing to solve it.

4. Aug 15, 2011

### J-Girl

Re: Inequalities- tricky question!!

ohh sorry i didnt even put the rest of the question in!! it was (x^2-4x+7)/(x^2+x-6)≤ 0
i just wasnt sure if i was allowed to automatically multiply by the denominator because i didnt know if it was positive or negative. if i rephrase it as (x-2)^2 + 3, i still cant cancel out any brackets because its not completely factored. sorry:( just havent done maths in ages and i suck at it..

5. Aug 15, 2011

### disregardthat

Re: Inequalities- tricky question!!

Can you tell when (x-2)^2+3 is positive and negative though? Notice the squared term.

6. Aug 29, 2011

### J-Girl

Re: Inequalities- tricky question!!

oohhh i got it now!!!:) yay took ages tho lol but thanks every1:)