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

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

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

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

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

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

1. Note that $x^2 -2x +3 = (x-1)^2 +2$, which is always positive. So you just need to consider the sign of the denominator, $x(x - 2)$, in the each of the ranges $x<0, 02$.

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

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

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

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