Solve (2x+1)^2-5xleq(2x+3)(2x-3)

$$( 2 x + 1 ) ^ { 2 } - 5 x \leq ( 2 x + 3 ) ( 2 x - 3 )$$

Solve for x

$$x\geq 10$$

Solution Steps

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

$$4x^{2}+4x+1-5x\leq \left(2x+3\right)\left(2x-3\right)$$

Combine 4x and -5x to get -x.

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

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.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

Subtract 4x^{2} from both sides.

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

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

$$-x+1\leq -9$$

Subtract 1 from both sides.

$$-x\leq -9-1$$

Subtract 1 from -9 to get -10.

$$-x\leq -10$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x\geq \frac{-10}{-1}$$

Fraction \frac{-10}{-1} can be simplified to 10 by removing the negative sign from both the numerator and the denominator.

$$x\geq 10$$