Rational equations and inequalities

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Equations and inequalities with Rational Functions

Rational equations and inequalities

To solve equations involving rational expressions, we have the freedom to clear out fractions before proceeding. After multiplying both sides by the common denominator, we are left with a polynomial equation.

Solve the equation $\frac {2}{x} + \frac {3x}{x+1} = 4.$

The common denominator is $x(x+1)$ . We multiply both sides by $x(x+1)$ to clear out the fractions. $\begin{align*} \frac{2}{x} + \frac{3x}{x+1} &= 4 \\ x(x+1) \left( \frac{2}{x} + \frac{3x}{x+1} \right) &= x(x+1) ( 4 )\\ x(x+1) \cdot \frac{2}{x} + x(x+1) \cdot \frac{3x}{x+1} &= 4x(x+1)\\ 2(x+1) + 3x(x) &= 4x^2 + 4x\\ 3x^2 + 2x + 2 &= 4x^2 + 4x\\ x^2 + 2x - 2 &= 0. \end{align*}$

The quadratic formula gives solutions as $\displaystyle x = \dfrac {-2 \pm \sqrt {12}}{2} = -1 \pm \sqrt {3}$ .

If we look back at the original equation, we notice that there are some numbers that we are not allowed to plug in for $x$ . When $x=0$ or $x=-1$ , the left-hand side of the equation is not defined due to a division by zero issue. Since neither $-1 + \sqrt {3}$ nor $-1-\sqrt {3}$ have such an issue, they are both solutions.

One solution of the equation $\dfrac {2}{x+1}+ \dfrac {1}{x+2} = 1$ is $x = \sqrt {3}$ . Find another solution. $x = \answer {-\sqrt {3}}$

Inequalities

When faced with nonlinear inequalities, such as those involving general rational functions, we make use of a sign chart. The inequality in the following example is not given in factored form, so we have some work to do.

Solve the inequality $\displaystyle x^2 + 5x \leq -10 -\dfrac {16}{x-2}$ .

We’ll begin by moving everything to one side, then combining them all together into a single fraction. $\begin{align*} x^2 + 5x &\leq -10 -\dfrac{16}{x-2}\\ x^2 + 5x +10 +\dfrac{16}{x-2} &\leq 0\\ \left(x^2+5x+10\right) \cdot \left(\dfrac{x-2}{x-2}\right) +\dfrac{16}{x-2} &\leq 0\\ \dfrac{x^3+3x^2-20}{x-2} + \dfrac{16}{x-2} &\leq 0\\ \dfrac{x^3+3x^2-4}{x-2} &\leq 0\\ \dfrac{(x-1)(x+2)^2}{x-2} &\leq 0 \end{align*}$

Now that the inequality is in a better form for us to work with, we’ll build a sign chart like we did in the last example.

We see from the chart that $\displaystyle \dfrac {(x-1)(x+2)^2}{x-2}$ will be negative in $(1,2)$ . At $x=-2$ and $x=1$ it is zero. The solution is then: $\left \{ -2\right \} \cup [ 1, 2 )$ .

Find the solution of the inequality: $\displaystyle x + \dfrac {9}{x-1} > 5$ .

$(1, \infty )$ $[1,\infty )$ $(1,2)\cup (2,\infty )$ $\{1\} \cup [2,\infty )$ none of the above

Solve the inequality $\frac {1}{x-2} - \frac {4}{x+1} \leq 3$

We’ll start by moving everything to the left-hand side and combining them into a single fraction. $\begin{align*} \frac{1}{x-2} - \frac{4}{x+1} &\leq 3 \\ \frac{1}{x-2} - \frac{4}{x+1} - 3 &\leq 0\\ \frac{1(x+1)}{(x-2)(x+1)} - \frac{4(x-2)}{(x+1)(x-2)} - \frac{3(x+1)(x-2)}{(x+1)(x-2)} &\leq 0\\ \frac{(x+1) - (4x-8) - (3x^2-3x-6)}{(x+1)(x-2)} &\leq 0\\ \frac{-3x^2+15}{(x+1)(x-2)} &\leq 0\\ \frac{-3(x^2-5)}{(x+1)(x-2)} &\leq 0 \end{align*}$

To solve this, we’ll construct a sign chart. We start by noticing that $x^2-5 =0$ only if $x = \pm \sqrt {5}$ .

We see from the sign chart, that the solution is $\displaystyle \left ( -\infty , -\sqrt {5} \right ] \bigcup \left (-1, 2\right ) \bigcup \left [ \sqrt {5}, \infty \right )$ .