Question 200576
Given equation: {{{abs(x^2 + 2abs(x) + 3) > 4}}}


Case 1: x > 0
{{{abs(x^2 + 2x + 3) > 4}}}
{{{abs((x+1)^2 + 2) > 4}}}
{{{(x+1)^2 + 2 > 4}}}
{{{(x+1)^2 + 2 - 4 > 4 - 4}}}
{{{(x+1)^2 - 2 > 0}}}
{{{(x+1)^2 - (sqrt(2))^2 > 0}}}
{{{((x+1) + sqrt(2))*((x+1) - sqrt(2)) > 0}}}
{{{(x+1+sqrt(2))*(x+1-sqrt(2)) > 0}}}
{{{x = sqrt(2) - 1}}} [since the other solution is not feasible as x > 0]


Case 2: x < 0
{{{abs(x^2 - 2x + 3) > 4}}}
{{{abs((x-1)^2 + 2) > 4}}}
{{{(x-1)^2 + 2 > 4}}}
{{{(x-1)^2 + 2 - 4 > 4 - 4}}}
{{{(x-1)^2 - 2 > 0}}}
{{{(x-1)^2 - (sqrt(2))^2 > 0}}}
{{{((x-1) + sqrt(2))*((x-1) - sqrt(2)) > 0}}}
{{{(x-1+sqrt(2))*(x-1-sqrt(2)) > 0}}}
{{{x = -sqrt(2) + 1}}} [since the other solution is not feasible as x < 0]