# Solving Rational Inequalities Using Method of Intervals

Solving inequalities of the form (p(x))/(q(x))>0 (instead of ">" there can be another symbol of inequality), where p(x) and q(x) are polynomials, is based on the following reasoning.

Consider function h(x)=((x-a)(x-b))/((x-c)(x-d)), where a<b<c<d. If x>d, then each of factors x-a,x-b,x-c,x-d is positive, therefore, on interval (d,+oo) we have that h(x)>0. If c<x<d, then x-d<0, and other factors are still positive. Therefore, on interval (c,d) we have that h(x)<0. Similarly on interval (b,c) we have that h(x)>0 etc. (whole process is shown on figure a)).

Change of signs of function h(x) is convenient to draw with the help of wave-like curve (it is called curve of signs), which we draw from right to left, starting from top (figure b)). This illustration tells us, that on those intervals, where this curve is above coordinate line, we have that h(x)>0, on those intervals where curve is below coordinate line, we have that h(x)<0.

For above reasoning it doesn't matter how many there are linear factors in numerator and denominator, and positional relationship of roots of numerator and denominator of fraction on coordinate line. THerefore, above reasoning is applicable to the functions of the form f(x)=((x-a_1)(x-a_2)...(x-a_n))/((x-b_1)(x-b_2)...(x-b_k)), where numbers a_1,a_2,a_3,...,a_n,b_1,b_2,...,b_k are pairwise different. Change of signs of the function f(x) is also illustrated with the help of the curve of signs, which we draw from the right to left, starting from top, and draw through all points a_1,a_2,...a_n,b_1,b_2,...,b_k. This reasoning is a base for the method of intervals, which is used for solving rational inequalities.

Example 1. Solve inequality ((x+5)(x-sqrt(3))(x+sqrt(2)))/((2x-3)(4x+5))<0.

We need first to transform inequality into standard form:

((x+5)(x-sqrt(3))(x+sqrt(2)))/(2(x-3/2)*4(x+5/4))<0 .

Now, mulitply both sides by 8:

((x+5)(x-sqrt(3))(x+sqrt(2)))/((x-3/2)(x+5/4))<0 , which is equivalent to the initial inequality.

Change of signs of the function f(x)=((x+5)(x-sqrt(3))(x+sqrt(2)))/((x-3/2)(x+5/4))<0 is illustrated using curve of signs. Values x, for which f(x)<0 (corresponding intervals are dashed), satisfy following inequalities: x<-5;-sqrt(2)<x<-5/4;3/2<x<sqrt(3). This is solution of initial inequality.

Example 2. Solve inequality ((x-3)(x+2))/(x^2-1)<=1.

We need first to transform inequality into standard form:

((x-3)(x+2))/(x^2-1)-1<=0 ;

(x^2-x-6)/(x^2-1)-1<=0 ;

(x^2-x-6)/(x^2-1)-(x^2-1)/(x^2-1)<=0 ;

(x^2-x-6-(x^2-1))/(x^2-1)<=0;

(-x-5)/((x-1)(x+1))<=0;

Now, mulitply both sides by -1 and change sign:

((x+5))/((x-1)(x+1)<=0 , which is equivalent to the initial inequality.

Change of signs of the function f(x)=(x+5)/((x-1)(x+1))<0 is illustrated using curve of signs. Values x, for which f(x)<=0 (corresponding intervals are dashed), satisfy following inequalities: -5<=x<-1;x>1. This is solution of initial inequality.

Note, that we have inequality <=0, therefore we included point 5 into answer (on figure it is solid point). We didn't include 1 and -1, because these points are not in domain of the function f(x) (denominator equals 0); these points are shown on the figure as hole points.