• May 12th 2009, 08:25 PM
SVXX
How would I solve the following quadratic inequalities?

1.$\displaystyle \frac{x^{2} - 2x + 3}{x^{2} - 2x} < 0$

2.$\displaystyle \frac{3x + 4}{x^{2} -3x + 2} \geq 0$
• May 12th 2009, 10:30 PM
Hello SVXX
Quote:

Originally Posted by SVXX
How would I solve the following quadratic inequalities?

1.$\displaystyle \frac{x^{2} - 2x + 3}{x^{2} - 2x} < 0$

2.$\displaystyle \frac{3x + 4}{x^{2} -3x + 2} \geq 0$

1. Note that $\displaystyle x^2 -2x +3 = (x-1)^2 +2$, which is always positive. So you just need to consider the sign of the denominator, $\displaystyle x(x - 2)$, in the each of the ranges $\displaystyle x<0, 0<x<2, x>2$.

2. $\displaystyle \frac{3x + 4}{x^{2} -3x + 2} = \frac{3x+4}{(x-1)(x-2)}\ge 0$

Consider the values of $\displaystyle x$ which make each term zero; these are $\displaystyle -\tfrac43, 1, 2$, and look at the sign of the expression as $\displaystyle x$ moves along the number line from left to right through these values; in other words in the ranges

• $\displaystyle x<-\tfrac43$
• $\displaystyle -\tfrac43 < x < 1$
• $\displaystyle 1<x<2$
• $\displaystyle x>2$

For instance, when $\displaystyle x<-\tfrac43$, you get $\displaystyle \frac{-^{ve}}{-^{ve}\times -^{ve}}=-^{ve}$. So $\displaystyle x<-\tfrac43$ is not part of the solution set.