Question

Solve for x:
|x+2-x^2| + |x+2| = x^2

Answer 1

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

Given equation is

$\rm :\longmapsto\: |x + 2 - {x}^{2} | + |x + 2| = {x}^{2}$

can be rewritten as

$\rm :\longmapsto\: | - ({x}^{2} - x - 2)| + |x + 2| = {x}^{2}$

$\rm :\longmapsto\: | - ({x}^{2} - 2x + x - 2)| + |x + 2| = {x}^{2}$

$\rm :\longmapsto\: | - (x(x - 2) + 1( x - 2))| + |x + 2| = {x}^{2}$

$\rm :\longmapsto\: | - (x + 1)(x - 2)| + |x + 2| = {x}^{2}$

$\rm :\longmapsto\: | (x + 1)(2 - x)| + |x + 2| = {x}^{2}$

So, breaking points are

$\rm :\longmapsto\: - 2, \: - 1, \: 2$

We know,

Definition of Modulus function is

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

So,

We take out the first case,

$\rm :\longmapsto\:When \: x \: \leqslant - 2$

So,

$\rm :\longmapsto\: | (x + 1)(2 - x)| + |x + 2| = {x}^{2}$

reduces to

$\rm :\longmapsto\: - (x + 1)(2 - x) - (x + 2)= {x}^{2}$

$\rm :\longmapsto\: (x + 1)(x - 2) - (x + 2)= {x}^{2}$

$\rm :\longmapsto\: {x}^{2} - x - 2 - x - 2 = {x}^{2}$

$\rm :\longmapsto\: - 2x - 4 = 0$

$\rm :\longmapsto\: - 2x = 4$

$\bf\implies \:x = - 2 - - - (1)$

Now, Case :- 2

$\rm :\longmapsto\:When - 2 < x < - 1$

So,

$\rm :\longmapsto\: | (x + 1)(2 - x)| + |x + 2| = {x}^{2}$

reduces to

$\rm :\longmapsto\: - (x + 1)(2 - x) + (x + 2)= {x}^{2}$

$\rm :\longmapsto\: (x + 1)(x - 2) + (x + 2)= {x}^{2}$

$\rm :\longmapsto\: {x}^{2} - x - 2 + x + 2 = {x}^{2}$

$\rm :\longmapsto\:0 = 0$

Hence,

Solution is

$\bf\implies \:x \: \in \: ( - 2, - 1) - - (2)$

Now, Case :- 3

$\rm :\longmapsto\:When - 1 \leqslant x < 2$

So,

$\rm :\longmapsto\: | (x + 1)(2 - x)| + |x + 2| = {x}^{2}$

reduces to

$\rm :\longmapsto\: (x + 1)(2 - x) + x + 2= {x}^{2}$

$\rm :\longmapsto\: x + 2 - {x}^{2} + x + 2= {x}^{2}$

$\rm :\longmapsto\: 2x + 4 - 2{x}^{2} = 0$

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

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

$\rm :\longmapsto\: x(x - 2) + 1(x - 2) = 0$

$\rm :\longmapsto\: (x - 2)(x + 1) = 0$

$\rm :\longmapsto\:x = 2 \: \: or \: \: x = - 1$

$\bf\implies \:x = - 1 - - - (3)$

Now, Case :- 4

$\rm :\longmapsto\:When \: x \: \geqslant \: 2$

So,

$\rm :\longmapsto\: | (x + 1)(2 - x)| + |x + 2| = {x}^{2}$

reduces to

$\rm :\longmapsto\: - (x + 1)(2 - x) + (x + 2)= {x}^{2}$

$\rm :\longmapsto\: (x + 1)(x - 2) + (x + 2)= {x}^{2}$

$\rm :\longmapsto\: {x}^{2} - x - 2 + x + 2 = {x}^{2}$

$\rm :\longmapsto\:0 = 0$

Hence, Solution is

$\bf\implies \:x \geqslant 2 - - - (4)$

So, on combining equation (1), (2), (3) and (4), we get

$\bf\implies \:x \: \in \: [ - 2, \: - 1] \: \cup \: [2, \: \infty )$