Solve x^2+x^2-2x+x^2+2x-3geq0

$${ x }^{ 2 } + { x }^{ 2 } -2x+ { x }^{ 2 } +2x-3 \geq 0$$

Solve for x

$$x\in (-\infty,-1]\cup [1,\infty)$$

Solution Steps

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-2x+x^{2}+2x-3\geq 0$$

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$3x^{2}-2x+2x-3\geq 0$$

Combine -2x and 2x to get 0.

$$3x^{2}-3\geq 0$$

Add 1 to both sides.

$$x^{2}\geq 1$$

Calculate the square root of 1 and get 1. Rewrite 1 as 1^{2}.

$$x^{2}\geq 1^{2}$$

Inequality holds for |x|\geq 1.

$$|x|\geq 1$$

Rewrite |x|\geq 1 as x\leq -1\text{; }x\geq 1.

$$x\leq -1\text{; }x\geq 1$$