Question

Solve the following inequality: \[\dfrac{{{x^2} - 9}}{{3x - {x^2} - 24}} < 0\].
A. \[3,3\]
B. \[3, - 3\]
C. Does not exists
D. Cannot be determined

Answer 1

Hint: The given problem revolves around the concept of algebraic solution to solve the equation to get the desired value (or, roots). As a result, by multiplying the whole equation by its denominator. Then, using an algebraic identity such as \[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\], the desired root is obtained.

Complete step-by-step solution:
Since, we have given that
The given inequality is,
$\dfrac{{{x^2} - 9}}{{3x - {x^2} - 24}} < 0$
Now, we have a quadratic expression $3x - {x^2} - 24$. Comparing it with the standard quadratic equation $a{x^2} + bx + c = 0$, we get, $a = -1$, $b = 3$ and $c = - 24$.
Now, calculating the discriminant of the expression using the formula $D = {b^2} - 4ac$ , we get,
$D = {3^2} - 4\left( { - 1} \right)\left( { - 24} \right)$
$ \Rightarrow D = 9 - 96 = - 87$
Since the discriminant is negative. So, the function has no real roots, which in turn means that the expression is either positive or negative for all real values of x.
This is decided by the opening of the parabola judged by the coefficient of ${x^2}$ of the quadratic expression. So, the coefficient of ${x^2}$ is $a = -1$. This means that the expression acquires negative values for each value of x.
Multiplying both sides of the above equation by $3x - {x^2} - 24$ , we will have to reverse the sign of inequality as the expression $3x - {x^2} - 24$.
Hence, we get,
$ \Rightarrow {x^2} - 9 > 0$
Using the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$, we get,
$ \Rightarrow \left( {x - 3} \right)\left( {x + 3} \right) > 0$
Now, since the product of two factors is positive, either both the terms should be positive or both terms should be negative.
So, the first solution set is: $\left( {x - 3} \right) > 0$ and $\left( {x + 3} \right) > 0$
$x > 3$ and $x > - 3$
Intersection of both the inequalities is: $x > 3$.
Also, second solution set is: $\left( {x - 3} \right) < 0$ and $\left( {x + 3} \right) < 0$
$x < 3$ and $x < - 3$
Intersection of both the inequalities is: $x < - 3$.
So, the final solution of the whole inequality is the union of both the solution sets. So, we have, $x \in \left( { - \infty, - 3} \right),\left( {3, \infty } \right)$.

Note: If a quadratic expression has a negative discriminant, then the expression has no real roots and is either positive or negative for all values of the variable. Reverse the sign of inequality if you are multiplying both the sides of an inequality by a negative number. The algebraic identities such as $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$ helps in factoring the expression and solving the inequality easily.