Solve 4x^2<(2x-1)^2

$$4 x ^ { 2 } < ( 2 x - 1 ) ^ { 2 }$$

Solve for x

$$x<\frac{1}{4}$$

Solution Steps

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

$$4x^{2}<4x^{2}-4x+1$$

Subtract 4x^{2} from both sides.

$$4x^{2}-4x^{2}<-4x+1$$

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

$$0<-4x+1$$

Swap sides so that all variable terms are on the left hand side. This changes the sign direction.

$$-4x+1>0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-4x>-1$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x<\frac{-1}{-4}$$

Fraction \frac{-1}{-4} can be simplified to \frac{1}{4} by removing the negative sign from both the numerator and the denominator.

$$x<\frac{1}{4}$$