Lesson Objective

Dear learner,

By the end of this lesson, you will be able to:

• Define rational equations.
• Solve problems given on rational equations.
• Compute the intervals where rational inequalities have possible solution.

Key Terms

• Rational Equation
• Rational Inequality
• Sign chart

Brainstorming Questions

a. What is a rational equation?

b. How do you identify a rational equation?

c. What is the difference between a rational expression and a rational equation?

2.1. Rational Equations

$\textbf{Definition:}$

An equation that can be written in the form of $ \frac{P(x)}{Q(x)} = 0 $, where $P(x) $ and $ Q(x) $ are polynomials, $ Q(x) \neq 0 $ is a rational equation.

Example 1:

Identify rational equations

a. $\frac{x^2 + 5 \ln x}{6x + 7} = 0 $
b. $3x^3 – 5x^2 + 6 = 2 – 5x^2 $
c. $\frac{5x – 6}{3^x} = 0 $
d. $ \frac{3 – x^{-2}}{x^{-3} + 1} = 0 $

Solution:

$\text{a.} \frac{x^{2}+5lnx}{6x+7}=0 \text{ is not a rational equation because}~x^{2}+5lnx~\text{is not polynomial.}$

$\text{b.} 3x^{3}-5x^{2}+6=2-5x^{2} \text{ is a rational equation.}$

$\text{c.} \frac{5x-6}{3^{x}}=0 \text{ is not a rational equation because}~3^{x}~\text{is not polynomial.}$

$\text{d.} \frac{3-x^{-2}}{x^{-2}+1}=0 \text{ is a rational equation, because} \frac{3-x^{-2}}{x^{-2}+1}=0 \text{ can be written}\frac{x(3x^{2}-1)}{1+x^{2}}=0$.

Solving Rational Equations:

Solving rational equations is finding all the possible solutions in the given domain of an expression.

$\text {To solve rational equations of the form}\frac{P(x)}{Q(x)}=0, \text {we just follow the steps below}$:

$\text{Step 1: restrict the domain, i.e}~Q(x)\neq0$

$\text{Step 2: solve the polynomial equation}~P(x)=0$

$\text{Step 3: put the solution set}$

Example 2:

Find the solution set of the following equations.

$\text{a.} \frac{x^{2}-5x+6}{x^{2}-2}=0$

$\text{b.} \frac{x}{x-1}=\frac{3}{x+1}$

$\text{c.} \frac{5}{x-1}+\frac{1}{4-3x}=\frac{3}{6x-8}$

$\text{d.} \frac{x^{2}}{x+3}+\frac{3x}{x^{2}+x-6}=\frac{2x}{x+3}$

$\text{e.} \frac{1+\frac{2}{x}}{x-\frac{1}{x}}=0$

$\text{f.} \frac{x+4}{x-5}-\frac{1}{x+5}=\frac{10}{x^{2}-25}$

Solution:

$\text{a.} \frac{x^{2}-5x+6}{x^{2}-2}=0$

$\text{Domain:}~x^{2}-2=0\Rightarrow x=\sqrt{2}~\text{or}~x=-\sqrt{2}$.

Thus, the domain is all real number except $x =\sqrt{2} \;and \; -\sqrt{2}$.

$\Rightarrow x^{2}-5x+6=0$

$\Rightarrow (x-2) (x-3)=0$

$\Rightarrow x=2~\text{or}~x=3$.

$\text{Therefore}~S.S=\{2,3\}$.

$\text{b.} \frac{x-1}{x+1}=\frac{2}{x+1}. \text{Domain=R \setminus {-1,1}}$

$\Rightarrow x(x+1)-x-1=2(x+1)$

$\Rightarrow x^{2}-1=2x-2$

$ \Rightarrow x^{2}-2x-3=0$

$ \Rightarrow (x-3)(x+1)=0$

$ \Rightarrow x=-1 \;or\; x=3$. But $-1$ is not in the domain.

$\text{Therefore}~S.S=\{3\}$.

$\text{c.} \frac{5}{x-1}+\frac{1}{4-3x}=\frac{3}{6x-8}$.

$\text{Take the denominators:}~x-1=0\Rightarrow x=1,~4-3x=0\Rightarrow x=\frac{4}{3}$.

Thus, the domain is $R\setminus \{1, \frac{4}{3}\}$. Now let us find the solution:

$\Rightarrow\frac{5(4-3x)+(x-1)}{(x-1)(4-3x)}=\frac{-3}{2(4-3x)}$

$\Rightarrow\frac{20-15x+x-1}{x-1}=-\frac{3}{2}$

$\Rightarrow2(19-14x)=3-3x$

$\Rightarrow38-28x=3-3x$

$\Rightarrow35=25x\Rightarrow x=\frac{7}{5}$.

$\text{Therefore}~S.S=\{\frac{7}{5}\}$.

$\text{d.} \frac{x^{2}}{x+3}+\frac{3x}{x^{2}+x-6}=\frac{2x}{x+3}$.

$\text{ Taking the denominators:}~x+3=0\Rightarrow x=-3,~x^{2}+x-6=0\Rightarrow x=2~\text{or}~x=-3$.

Thus, the domain is $R\setminus \{-3,2\}.$ Now let us find the solution:

$\Rightarrow\frac{x^{2}(x-2)+3x}{(x-2)(x+3)}=\frac{2x(x-2)}{(x-2)(x+3)}$

$\text{LCM}=(x-2)(x+3)$

$\Rightarrow x^{2}(x-2)+3x=2x(x-2)$

$\Rightarrow x^{3}-2x^{2}+3x=2x^{2}-4x$

$\Rightarrow x^{3}-4x^{2}+7x=0$

$\Rightarrow x(x^{2}-4x+7)=0$

$\Rightarrow x=0~\text{or}~x^{2}-4x+7=0$

$\Rightarrow x=0~\text{or}~x^{2}-4x+7\neq0$.

Thus, $s.s=\{0\}$.

$\text{e.} \frac{1+\frac{2}{x}}{x-\frac{1}{x}}=0. \text{Domain:}~x\neq -1,0,1$. Now let us find the solution.

$\Rightarrow 1+\frac{2}{x}=0$

$\Rightarrow \frac{2}{x}=-1$

$\Rightarrow-x=2$

$\Rightarrow x=-2$.

$\text{Therefore}~S.S=\{-2\}$.

$\text{f.} \frac{x+4}{x-5}-\frac{1}{x+5}=\frac{10}{x^{2}-25}. \text{Domain:}~x\neq-5,5$. Let us solve the equation.

$\Rightarrow\frac{(x+4)(x+5)-(x-5)}{(x-5)(x+5)}=\frac{10}{(x-5)(x+5)}$

$\Rightarrow(x+4)(x+5)-(x-5)=10$

$\Rightarrow x^{2}+9x+20-x+5-10=0$

$\Rightarrow x^{2}+8x+15=0$

$\Rightarrow(x+3)(x+5)=0$

$\Rightarrow x+3=0~\text{or}~x+5=0$

$\Rightarrow x=-3~\text{or}~x=-5$. But $-5$ is not in the domain.

$\text{Therefore}~ S.S=\{-3\}$.

2.2. Rational Inequalities

Definition

A rational inequality is an inequality which can be written of the form:

$\frac{p(x)}{q(x)} \leq 0 \text{or} \frac{p(x)}{q(x)} \geq 0$, Where $p(x) $ and $ q(x) $ are polynomial expressions, $ q(x) \neq 0 $.

Example 3

Which of the following are rational inequalities?

$\text{a.} \frac{35x+6}{6-25x}<25-14x$

$\text{b.} 3^{x}>\frac{58-x}{5x}$

$\text{c.} \frac{5x+\sin{x}}{8x+6}\ge0$

$\text{d.} \frac{6-21x^{2}+1}{x^{2}+5x+6}\ge0$.

Solution:

$\text{a.} \frac{35x+6}{6-25x}<25-14x$ and $\text{d.} \frac{6-21x^{2}+1}{x^{2}+5x+6}\ge0$ are rational inequalities.

$\text{b.} 3^{x}>\frac{58-x}{5x}$ is not rational inequality because of $3^x$.

$\text{c.} \frac{5x+\sin{x}}{8x+6}\ge0$ is not rational inequality because of $5x+sinx$.

Solving Rational Inequalities

Rational inequalities involving polynomial expressions are solved in various techniques but the very basic one is applying the principle of multiplication of real system.

Note:

$\text{Let}~\frac{P(x)}{Q(x)}~\text{be a rational expression then}$

$\text{Case 1. If}~\frac{P(x)}{Q(x)}>0,~\text{then}~P(x)~\text{and}~Q(x)~\text{have the same sign}$.

$\text{i.e.}~P(x)>0~\text{and}~Q(x)>0~\text{or}~P(x)<0~\text{and}~Q(x)<0$.

$\text{Case 2. If}~\frac{P(x)}{Q(x)}<0,~\text{then}~P(x)~\text{and}~Q(x)~\text{have opposite sign}$.

$\text{i.e.}~P(x)>0~\text{and}~Q(x)<0~\text{or}~P(x)<0~\text{and}~Q(x)>0$

Example 4:

Solve $\frac{x+2}{x-1} < 0$

Solution:

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

$\text{i.}~x+2>0~\text{and}~x-1<0\Rightarrow x>-2~\text{and}~x<1$

$\Rightarrow x>-2\cap x<1=-2<x<1$

$\Rightarrow S.S_{1}=-2<x<1=(-2,1)$

$\text{ii.}~x+2<0~\text{and}~x-1>0\Rightarrow x<-2~\text{and}~x>1$

$\Rightarrow x<-2\cap x>1={} \Rightarrow S.S_{2}=\emptyset$.

$\text{Therefore the solution set becomes}$

$\Rightarrow S.S=S.S_{1}\cup S.S_{2}=(-2,1)\cup\emptyset=(-2,1)$.

General Steps Solving Rational Inequalities Using Sign Chart

Solving rational inequalities using sign chart is based on the principle of multiplication of real system. It is directedly derived from the above basic principle combined with number lines to show intervals where the possible solution is obtained.

$\text{General steps solving rational inequalities using sign chart}$.

$\text{Step 1: Restrict the domain}$.

$\text{Step 2: Factorize the numerators and denominators}$.

$\text{Step 3: Determine the boundary points on a number line to get intervals on the sign chart}$.

$\text{Step 4: Check the sign of each factor on each boundary}$.

$\text{Step 5: Put the solution according to the sign of inequality by observing the sign if last row}$.

Example 5

$\text{a.} \frac{x+2}{x-1}\ge x+2$

$\text{b.} \frac{2x^{2}+x-1}{x^{2}-4x+4}\ge0$

$\text{c.} \frac{1}{x}\le\frac{x}{x-1}+1$

$\text{d.} \frac{x^{2}}{x^{2}+2}\ge0$

Solution:

$\text{a.} \frac{x+2}{x-1}\ge x+2$

$\text{Step 1: Domain}=\mathbb{R}/\{1\}$

$\text{Re-arrange the inequality}$, we have

$\frac{x+2}{x-1}-(x+2)\ge0$

$\Rightarrow\frac{x+2-(x+2)(x-1)}{x-1}\ge0$

$\Rightarrow\frac{x+2-(x^{2}-x+2x-2)}{x-1}\ge0$

$\Rightarrow\frac{x+2-x^{2}-x+2}{x-1}\ge0$

$\text{Step 2:} \frac{x^{2}-4}{x-1}\le0 \Leftrightarrow\frac{(x-2)(x+2)}{x-1}=0$

$\text{Step 3: The boundary points are}~x=-2, 1~\text{and}~2$

$\text{Step 4: Construct sign chart}$.

The inequality is no existed at $x = 1$ symbol in the table shows the set does not include that point and the Solution set contains the intervals with negative sign on the last row. Therefore, the solution set is given as follows:

$S.S = (-\infty,-2] \bigcup (1, 2]$.

$\text{b.} \frac{2x^{2}+x-1}{x^{2}-4x+4}\ge0 \Leftrightarrow\frac{(2x-1)(x+1)}{(x-2)^{2}}\ge0.$

$\text{Domain}=\mathbb{R}/\{2\}$.

$\text{Boundary points are}~x=-1,\frac{1}{2}~\text{and}~2$.

Hence, we have the sign chart below:

Therefore, the solution is given by $S.S=(-\infty,-1]\cup[\frac{1}{2},2)\cup(2,\infty)$.