Solve left(2x-3right)^2<4(x-1)(1+x)

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

Solve for x

$$x>\frac{13}{12}$$

Solution Steps

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

$$4x^{2}-12x+9<4\left(x-1\right)\left(1+x\right)$$

Use the distributive property to multiply 4 by x-1.

$$4x^{2}-12x+9<\left(4x-4\right)\left(1+x\right)$$

Use the distributive property to multiply 4x-4 by 1+x and combine like terms.

$$4x^{2}-12x+9<4x^{2}-4$$

Subtract 4x^{2} from both sides.

$$4x^{2}-12x+9-4x^{2}<-4$$

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

$$-12x+9<-4$$

Subtract 9 from both sides.

$$-12x<-4-9$$

Subtract 9 from -4 to get -13.

$$-12x<-13$$

Divide both sides by -12. Since -12 is negative, the inequality direction is changed.

$$x>\frac{-13}{-12}$$

Fraction \frac{-13}{-12} can be simplified to \frac{13}{12} by removing the negative sign from both the numerator and the denominator.

$$x>\frac{13}{12}$$