# How to solve? x^2-4x+3 ________ <0 root of 2+x-x^2I know how to solve in 2 ways but I don't know which one is right? Do we need to put this 2+x-x^2 in table together with...

How to solve? x^2-4x+3

________ <0

root of 2+x-x^2

I know how to solve in 2 ways but I don't know which one is right?

Do we need to put this 2+x-x^2 in table together with x^2-4x+3,and then find solution, or we don't look that at all,just condition(1,2), because it could never be <0 ? Thanks in advance :)

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First we can factor into

(x - 3)(x - 1)/sqrt((2-x)(1+x)) < 0

The sqrt is by definition positive so we have to check when (x - 3)(x - 1) < 0

We have two cases

x - 3 < 0 and x - 1 > 0 this gives us 1 < x < 3

x - 3 > 0 and x - 1 < 0 this gives x < 1 and x > 3 there is no solution to this.

The only other condition is (2-x)(1+x) > 0

We again have two cases because the argument to square root must be positive

2 - x > 0 and 1 + x > 0 this gives us x < 2 and x > -1 which means -1 < x < 2.

2 - x < 0 and 1 + x < 0 this gives us x > 2 and x < -1 which has no solution

Combining this with our previous result that 1 < x < 3 and -1 < x < 2

**we get the solution 1 < x < 2. or x is an element of (1, 2)**