Solve (2x+3)(2x-3)<4(x-2)(x+3)

$$( 2 x + 3 ) ( 2 x - 3 ) < 4 ( x - 2 ) ( x + 3 )$$

Solve for x

$$x>\frac{15}{4}$$

Solution Steps

Consider \left(2x+3\right)\left(2x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$\left(2x\right)^{2}-9<4\left(x-2\right)\left(x+3\right)$$

Expand \left(2x\right)^{2}.

$$2^{2}x^{2}-9<4\left(x-2\right)\left(x+3\right)$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-9<4\left(x-2\right)\left(x+3\right)$$

Use the distributive property to multiply 4 by x-2.

$$4x^{2}-9<\left(4x-8\right)\left(x+3\right)$$

Use the distributive property to multiply 4x-8 by x+3 and combine like terms.

$$4x^{2}-9<4x^{2}+4x-24$$

Subtract 4x^{2} from both sides.

$$4x^{2}-9-4x^{2}<4x-24$$

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

$$-9<4x-24$$

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

$$4x-24>-9$$

Add 24 to both sides.

$$4x>-9+24$$

Add -9 and 24 to get 15.

$$4x>15$$

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

$$x>\frac{15}{4}$$