How do you solve #(x^2-x-2)/(x^2-4x+3)>0#?

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Answer 1 · 2016-07-11T21:41:53

#x in (-oo, -1) uu (1, 2) uu (3, oo)#

#(x^2-x-2)/(x^2-4x+3) = ((x-2)(x+1))/((x-1)(x-3))#

So the numerator changes sign at #x=-1# and #x=2# while the denominator changes sign at #x=1# and #x=3#

Each change of sign (numerator or denominator) changes the sign
of the quotient.

So the sign reverses each time we move to the next interval in the sequence:

#(-oo, -1), (-1, 1), (1, 2), (2, 3), (3, oo)#

For large positive values of #x#, all of the linear factors are positive, so #(x^2 - x - 2)/(x^2-4x+3) > 0#.

So the signs of the quotient in the #5# intervals listed are:

#+ - + - +#

So the quotient is positive in #(-oo, -1) uu (1, 2) uu (3, oo)#

graph{(x^2-x-2)/(x^2-4x+3) [-10, 10, -5, 5]}