Solve 4x^2+2xgeq2x+16

$$4 x ^ { 2 } + 2 x \geq 2 x + 16$$

Solve for x

$$x\in (-\infty,-2]\cup [2,\infty)$$

Solution Steps

Subtract 2x from both sides.

$$4x^{2}+2x-2x\geq 16$$

Combine 2x and -2x to get 0.

$$4x^{2}\geq 16$$

Divide both sides by 4. Since 4 is positive, the inequality direction remains the same.

$$x^{2}\geq \frac{16}{4}$$

Divide 16 by 4 to get 4.

$$x^{2}\geq 4$$

Calculate the square root of 4 and get 2. Rewrite 4 as 2^{2}.

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

Inequality holds for |x|\geq 2.

$$|x|\geq 2$$

Rewrite |x|\geq 2 as x\leq -2\text{; }x\geq 2.

$$x\leq -2\text{; }x\geq 2$$