We offer top techniques to solve rational inequalities for high school students. It is an important part of algebra and calculus. Of course, there is a lot of methods to solve these inequalities. Some of them are based on advanced tools in analysis while others are mainly based on the simplification of the expression of the rational equation.

### Basics of rational fractions

Before talking about rational inequalities, it is better to first master the notion of calculus for rational numbers and fractions.A rational number is a real number that can be written as $x=\frac{a}{b}$, where $a$ and $b$ are renumbers with $b$ different of zero ($b\neq 0$). The number $a$ is called the

**numerator**and $b$ is the**denominator.**Then**a fraction is (numerator/denominator).**For details on computation for this class of numbers, we refer to the fractions chapter.

### Sign of a fraction

**Positive fraction:**The fraction $\frac{a}{b}$ is positive if its numerator and denominator have the same sign (positive together of negative together). This means that $a$ and $b$ are positive, or $a$ and $b$ are negative.

**Negative fraction:**The fraction $\frac{a}{b}$ is negative if the numerator and denominator have opposite signs. This means that $a$ is positive and $b$ is negative or $a$ is negative and $b$ is positive.

### The first order rational functions

In this paragraph, we will give you a unified and simple technique to solve rational inequalities of the forme

\begin{align}\frac{f(x)}{g(x)}\ge 0,\end{align}

where $f(x)$ and $g(x)$ are afine functions of the forme $ax+b$.

**Example I:**Solve the rational inequality

\begin{align}\frac{x-2}{x+1}\ge 0,\end{align}

Solution: As the fraction is postive then we have ($x-2\ge 0$ and $x+1>0$) or ($x-2\le0$ and $x+1<0$). This mean that \begin{align}x\ge 2\quad\text{and}\quad x>-1\end{align}

or

\begin{align}x\le 2\quad\text{and}\quad x<-1\end{align}

This means that $x\in [2,+\infty)$ or $x\in (-\infty,-1)$. Thus the set $\mathcal{S}$ of solutions of the rational inequality is

\begin{align} \mathcal{S}=[2,+\infty)\cup (-\infty,-1)\end{align}

**Example II:**Determine the solution of the following rational inequalilty

\begin{align} \frac{3x-2}{x+1}\ge 2.\end{align}

This problem can also be reformulated as

\begin{align} \frac{3x-2}{x+1}- 2\ge 0.\end{align}

By calculating this difference, we only need to solve the following rational inequality

\begin{align} \frac{3x-2}{x+1}- 2=\frac{x-3}{x+1}\ge 0.\end{align}

Using the same arguments as in Example I, we deduce that the set of solution is

\begin{align} \mathcal{S}=[3,+\infty)\cup (-\infty,-1).\end{align}

### Rational inequalities with quadratic functions

Determine the set of solutions $\mathcal{S}$ of the following rational inequality

\begin{align}\frac{3x^2+5x+2}{x^2+2x}\ge 0.\end{align}

The same principle can also be applied to these types of inequalities. As the solutions depend on the sign of the quadratic function $3x^2+5x+2$ and $x^2+2x$, then the first work to do is to determine signs of these functions. Using the chapter of quadratic equations, we can write $ 3x^2+5x+2=(x+1)(x+\frac{2}{3})$ and $ x^2+2x=x(x+2)$. Using the following table

we deduce that $3x^2+5x+2$ is positive if $x\in (-\infty,-1]\cup [-\frac{2}{3},+\infty)$, and it is negative if $x\in [-1,-\frac{2}{3}]$. We also have the table

Thus, $x^2+2x$ is strictly postitive if $x\in (-\infty,-2)\cup (0,+\infty)$, and it is negative if $x\in (-2,0)$.

Now $x$ is a solution of the rational inequality if $x\in (-\infty,-2)\cup [-1,-\frac{2}{3}]$.

**Note:**The same technique can also be applied for rational inequalities defined by the polynomial functions with higher degrees.

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