Question

Solve: x^2 - | x | - 12 < 0​

Answer 1

$step \: 1 \:$

$add \: 12 \: to \: both \: sides.$

${x}^{2} - |x| - 12 + 12 < 0 + 12$

${x}^{2} - |x| = < 12$

$step \: 2$

$Rewrite \: the \: inequality \: \\ without \: the \: absolute \: value.$

$12 < \:x < \: 12$

Answer 2

$\large\underline{\sf{Solution-}}$

Using definition of |x|, we have

$\begin{gathered}\begin{gathered}\bf\: |x| = \begin{cases} &\sf{ - x \: \: \: when \: x < 0} \\ &\sf{x \: \: \: whenx \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

So,

Given inequality is

$\rm :\longmapsto\: {x}^{2} - |x| - 12 < 0$

Two cases arises

Case :- 1

• When x < 0, we have |x| = - x

So,

$\rm :\longmapsto\: {x}^{2} - |x| - 12 < 0$

can be rewritten as

$\rm :\longmapsto\: {x}^{2} - ( - x) - 12 < 0$

$\rm :\longmapsto\: {x}^{2} + x - 12 < 0$

$\rm :\longmapsto\: {x}^{2} + 4x - 3x- 12 < 0$

$\rm :\longmapsto\:x(x + 4) - 3(x + 4) < 0$

$\rm :\longmapsto\:(x + 4)(x - 3) < 0$

$\bf\implies \: - 4 < x < 3$

$\rm :\longmapsto\:As \: x < 0$

$\bf\implies \:x \: \in \: ( - 4, \: 0) - - - (1)$

Case :- 2

$\rm :\longmapsto\:When \: x \geqslant 0, \: we \: have \: |x| = x$

So,

$\rm :\longmapsto\: {x}^{2} - |x| - 12 < 0$

can be rewritten as

$\rm :\longmapsto\: {x}^{2} - (x) - 12 < 0$

$\rm :\longmapsto\: {x}^{2} - x - 12 < 0$

$\rm :\longmapsto\: {x}^{2} - 4x + 3x - 12 < 0$

$\rm :\longmapsto\:x(x - 4) + 3(x - 4) < 0$

$\rm :\longmapsto\:(x - 4)(x + 3) < 0$

$\bf\implies \: - 3 < x < 4$

$\rm :\longmapsto\:As \: x \geqslant 0$

$\bf\implies \:x \: \in \: [0, 4) - - - (1)$

So from equation (1) and (2), we concluded that

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \: x \: \in \: ( - 4, \: 4)}$

Additional Information :-

$\rm :\longmapsto\:When \: \: a < b \: then$

$\sf \: (x - a)(x - b) < 0 \implies \: a < x < b$

$\sf \: (x - a)(x - b) \leqslant 0 \implies \: a \leqslant x \leqslant b$

$\sf \: (x - a)(x - b) \geqslant 0 \implies \: x \leqslant a \: \: or \: \: x \geqslant b$

$\sf \: (x - a)(x - b) > 0 \implies \: x < a \: \: or \: \: x > b$