Question

2|x+1|^(2)-|x+1|=3, then x=

Answer 1

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

Given equation is

$\rm :\longmapsto\:2 { |x + 1| }^{2} - |x + 1| = 3$

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

So, given equation can be rewritten as

$\rm :\longmapsto\:2 {(x + 1)}^{2} - |x + 1| - 3 = 0 - - - (1)$

Now, we know,

By definition of Modulus function,

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

Thus,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |x + 1| = \begin{cases} &\sf{ - (x + 1)\: \: if \: \: x < - 1} \\ &\sf{x + 1 \: \: if \: \: x \geqslant - 1} \end{cases}\end{gathered}\end{gathered}$

So, Two case arises

Case :- 1

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

$\rm :\implies\: |x + 1| = - (x + 1)$

So, equation (1) reduces to

$\rm :\longmapsto\:2 {(x + 1)}^{2} +( x + 1)- 3 = 0$

$\rm :\longmapsto\:2( {x}^{2} + 2x + 1) + x + 1 - 3 = 0$

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

$\rm :\longmapsto\:2{x}^{2} + 5x = 0$

$\rm :\longmapsto\:x(2x + 5) = 0$

$\rm :\implies\:x = 0 \: \: \: or \: \: \: - \dfrac{5}{2}$

$\rm :\longmapsto\:As \: x \: < \: - \: 1$

$\bf\implies \:x = - \: \dfrac{5}{2}$

Case :- 2

$\rm :\longmapsto\:When \: x \: \geqslant \: - \: 1$

$\rm :\implies\: |x + 1| = x + 1$

So, equation (1) can be rewritten as

$\rm :\longmapsto\:2 {(x + 1)}^{2} - ( x + 1)- 3 = 0$

$\rm :\longmapsto\:2( {x}^{2} + 2x + 1) - x - 1 - 3 = 0$

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

$\rm :\longmapsto\: {2x}^{2} + 3x - 2 = 0$

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

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

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

$\rm :\implies\:x = - 2 \: \: \: \: or \: \: \: \: x = \dfrac{1}{2}$

$\rm :\longmapsto\:As \: x \: \geqslant \: - \: 1$

$\bf\implies \:x =\: \dfrac{1}{2}$

Hence,

$\rm :\longmapsto\:2 { |x + 1| }^{2} - |x + 1| = 3 \: have \: solution \:$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\bf\ \:x =\: \dfrac{1}{2} \: \: or \: \: - \: \dfrac{5}{2} }$