Question

How do you solve $\dfrac{{{x^2} + x - 6}}{{{x^2} - 3x - 4}} \leqslant 0$ ?

Answer 1

Hint: We have been given to solve an inequation. The expression on the LHS is the ratio of two quadratic expressions. Solving this inequation means finding the values of $x$ for which the given inequation holds true. We will have to analyse different conditions for the given constraint to get the result. We will get a range of values as a result.

Complete step by step solution:
We have been given to solve the inequation $\dfrac{{{x^2} + x - 6}}{{{x^2} - 3x - 4}} \leqslant 0$.
We have to find the values of $x$ for which this inequation holds true.
To find the solution we have to break the given constraint of inequation into different conditions.
Case 1: For $\dfrac{{{x^2} + x - 6}}{{{x^2} - 3x - 4}} \leqslant 0$, $\left( {{x^2} + x - 6} \right) \leqslant 0$ and $\left( {{x^2} - 3x - 4} \right) > 0$
Case 2: For $\dfrac{{{x^2} + x - 6}}{{{x^2} - 3x - 4}} \leqslant 0$, $\left( {{x^2} + x - 6} \right) \geqslant 0$ and $\left( {{x^2} - 3x - 4} \right) < 0$
First we will solve for Case 1. We will find the range of $x$ which satisfies both $\left( {{x^2} + x - 6} \right) \leqslant 0$ and $\left( {{x^2} - 3x - 4} \right) > 0$.
$
  \left( {{x^2} + x - 6} \right) \leqslant 0 \\
   \Rightarrow \left( {{x^2} + x + \dfrac{1}{4} - 6 - \dfrac{1}{4}} \right) \leqslant 0 \\
   \Rightarrow \left( {{x^2} + x + \dfrac{1}{4} - \dfrac{{25}}{4}} \right) \leqslant 0 \\
   \Rightarrow {\left( {x + \dfrac{1}{2}} \right)^2} \leqslant \dfrac{{25}}{4} \\
   \Rightarrow - \dfrac{5}{2} \leqslant \left( {x + \dfrac{1}{2}} \right) \leqslant \dfrac{5}{2} \\
   \Rightarrow - \dfrac{5}{2} - \dfrac{1}{2} \leqslant x \leqslant \dfrac{5}{2} - \dfrac{1}{2} \\
   \Rightarrow - 3 \leqslant x \leqslant 2 \\
 $
We get, $x \in \left[ { - 3,2} \right]$
And,
$
\left( {{x^2} - 3x - 4} \right) > 0 \\
\Rightarrow \left( {{x^2} - 3x + \dfrac{9}{4} - 4 - \dfrac{9}{4}} \right) > 0 \\
\Rightarrow \left( {{x^2} - 3x + \dfrac{9}{4} - \dfrac{{25}}{4}} \right) > 0 \\
\Rightarrow {\left( {x - \dfrac{3}{2}} \right)^2} > \dfrac{{25}}{4} \\
\Rightarrow \left( {x - \dfrac{3}{2}} \right) > \dfrac{5}{2}\;\;or\;\;\left( {x - \dfrac{3}{2}} \right) < - \dfrac{5}{2} \\
\Rightarrow x > \dfrac{5}{2} + \dfrac{3}{2}\;\;or\;\;x < - \dfrac{5}{2} + \dfrac{3}{2} \\
\Rightarrow x > 4\;\;or\;\;x < - 1 \\
$
We get, $x \in \left( { - \infty , - 1} \right) \cup \left( {4,\infty } \right)$
We have to find the intersecting range for case 1.
$
x \in \left[ { - 3,2} \right] \cap \left\{ {\left( { - \infty , - 1} \right) \cup \left( {4,\infty } \right)} \right\} \\
\Rightarrow x \in \left[ { - 3, - 1} \right) \\
$
Now we try to solve the inequation using case 2. We will find the range of $x$ which satisfies both $\left( {{x^2} + x - 6} \right) \geqslant 0$ and $\left( {{x^2} - 3x - 4} \right) < 0$.
$
  \left( {{x^2} + x - 6} \right) \geqslant 0 \\
   \Rightarrow \left( {{x^2} + x + \dfrac{1}{4} - 6 - \dfrac{1}{4}} \right) \geqslant 0 \\
   \Rightarrow \left( {{x^2} + x + \dfrac{1}{4} - \dfrac{{25}}{4}} \right) \geqslant 0 \\
   \Rightarrow \left( {x + \dfrac{1}{2}} \right) \geqslant \dfrac{5}{2}\;\;or\;\;\left( {x + \dfrac{1}{2}} \right) \leqslant - \dfrac{5}{2} \\
   \Rightarrow x \geqslant \dfrac{5}{2} - \dfrac{1}{2}\;\;or\;\;x \leqslant - \dfrac{5}{2} - \dfrac{1}{2} \\
   \Rightarrow x \geqslant 2\;\;or\;\;x \leqslant - 3 \\
 $
We get, $x \in \left( { - \infty , - 3} \right] \cup \left[ {2,\infty } \right)$
And,
$
  \left( {{x^2} - 3x - 4} \right) < 0 \\
   \Rightarrow \left( {{x^2} - 3x + \dfrac{9}{4} - 4 - \dfrac{9}{4}} \right) < 0 \\
   \Rightarrow \left( {{x^2} - 3x + \dfrac{9}{4} - \dfrac{{25}}{4}} \right) < 0 \\
   \Rightarrow {\left( {x - \dfrac{3}{2}} \right)^2} < \dfrac{{25}}{4} \\
   \Rightarrow - \dfrac{5}{2} < \left( {x - \dfrac{3}{2}} \right) < \dfrac{5}{2} \\
   \Rightarrow - \dfrac{5}{2} + \dfrac{3}{2} < x < \dfrac{5}{2} + \dfrac{3}{2} \\
   \Rightarrow - 1 < x < 4 \\
 $
We get, $x \in \left( { - 1,4} \right)$
We have to find the intersecting range for case 2.
$
  x \in \left\{ {\left( { - \infty , - 3} \right] \cup \left[ {2,\infty } \right)} \right\} \cap \left( { - 1,4} \right) \\
   \Rightarrow x \in \left[ {2,4} \right) \\
 $
We got two different ranges of values for both the cases. The final result would be union of both the ranges as either of the conditions satisfied would also satisfy the given inequation.
$x \in \left[ { - 3, - 1} \right) \cup \left[ {2,4} \right)$
Thus, $x$ can take any value in this range to satisfy the given inequation.

Hence, the resulting solution for the given inequation is $x \in \left[ { - 3, - 1} \right) \cup \left[ {2,4} \right)$.

Note: We broke the given constraint of inequation into different conditions and solved the cases separately. The final result was obtained by calculating the union of the range of the values in different cases. We use intersection of range when both the conditions are required to be satisfied, and we use union of range when either of the conditions can be satisfied. To check the result we can use any value in the final range and put the value in the inequation to compare LHS and RHS.