How to solve? x^2-4x+3
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)
Sorry it doesn't look how it supose.
it is x^2-4x+3 / root of 2+x-x^2 <0
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