Question

Find the set of solutions of ${x^2} + x - 12 \leqslant 0$ by algebraic method.

Answer 1

Hint: We can rewrite the inequality by changing the $x$ term. $x$ can be replaced as $4x - 3x$. Then we can take the common factors out and simplify. This gives the solution set.

Formula used: Product of two terms equal to zero gives either one of the terms or both are zero.
Product of two terms is negative means exactly one of them is negative and the other is positive.

Complete step-by-step answer:
Given the inequality,
${x^2} + x - 12 \leqslant 0$
We are asked to find the set of solutions.
${x^2} + x - 12 \leqslant 0$
We can rewrite it as,
${x^2} + 4x - 3x - 12 \leqslant 0$
Taking $x$ from first two terms and $ - 3$ from last two terms common, we get
$ \Rightarrow x(x + 4) - 3(x + 4) \leqslant 0$
Again taking $x + 4$ as common we have,
$ \Rightarrow (x + 4)(x - 3) \leqslant 0$
This gives $(x + 4)(x - 3) = 0$ or $(x + 4)(x - 3) < 0$
So we have two cases.
We can consider each one separately.
Case (i):$(x + 4)(x - 3) = 0$
Product of two terms equal to zero gives either one of the terms or both are zero.
$ \Rightarrow x + 4 = 0{\text{ or }}x - 3 = 0$
We can subtract $4$ from first and add $3$ to the second inequality.
So we get,
$ \Rightarrow x = - 4{\text{ or }}x = 3$
Case (ii): $(x + 4)(x - 3) < 0$
Product of two terms is negative means exactly one of them is negative and the other is positive.
So we have, $x + 4 < 0{\text{ or }}x - 3 < 0$
We can subtract $4$ from first and add $3$ to the second inequality.
This gives, $x < - 4{\text{ or }}x < 3$
But this cannot happen simultaneously since the product will be positive.
Therefore $x < - 4{\text{ and }}x > 3$ and also $x > - 4{\text{ and }}x < 3$ are the chances.
But we can see that the first case $x < - 4{\text{ and }}x > 3$ does not happen. A number cannot be lesser than $ - 4$ and greater than $3$ at a time.
That is $x > - 4{\text{ and }}x < 3$
So the values that can be taken by $x$ according to this inequality is,
$x = - 4{\text{ , }}x = 3$ or $x > - 4{\text{ and }}x < 3$
That is $x \in [ - 4,3]$, closed interval.

$\therefore $The solution set of the given inequality is the set of all real numbers greater than or equal to $ - 4$ and less than or equal to $3$.

Note: By simplification of the inequality we can see that the values that can be taken by $x$ is the set of all real numbers greater than or equal to $ - 4$ and less than or equal to $3$. We can also solve the problem in another way. We can rewrite the inequality as $x(x + 1) \leqslant 12$. Solving we get the values.