Solve (2x-1)^2geq(2x+3)^2

$$( 2 x - 1 ) ^ { 2 } \geq ( 2 x + 3 ) ^ { 2 }$$

Solve for x

$$x\leq -\frac{1}{2}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$4x^{2}-4x+1\geq \left(2x+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$4x^{2}-4x+1\geq 4x^{2}+12x+9$$

Subtract 4x^{2} from both sides.

$$4x^{2}-4x+1-4x^{2}\geq 12x+9$$

Combine 4x^{2} and -4x^{2} to get 0.

$$-4x+1\geq 12x+9$$

Subtract 12x from both sides.

$$-4x+1-12x\geq 9$$

Combine -4x and -12x to get -16x.

$$-16x+1\geq 9$$

Subtract 1 from both sides.

$$-16x\geq 9-1$$

Subtract 1 from 9 to get 8.

$$-16x\geq 8$$

Divide both sides by -16. Since -16 is negative, the inequality direction is changed.

$$x\leq \frac{8}{-16}$$

Reduce the fraction \frac{8}{-16} to lowest terms by extracting and canceling out 8.

$$x\leq -\frac{1}{2}$$