Solve (3x-2)^2-(3x+4)(3x-4)geq8

$$( 3 x - 2 ) ^ { 2 } - ( 3 x + 4 ) ( 3 x - 4 ) \geq 8$$

Solve for x

$$x\leq 1$$

Solution Steps

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

$$9x^{2}-12x+4-\left(3x+4\right)\left(3x-4\right)\geq 8$$

Consider \left(3x+4\right)\left(3x-4\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 4.

$$9x^{2}-12x+4-\left(\left(3x\right)^{2}-16\right)\geq 8$$

Expand \left(3x\right)^{2}.

$$9x^{2}-12x+4-\left(3^{2}x^{2}-16\right)\geq 8$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-12x+4-\left(9x^{2}-16\right)\geq 8$$

To find the opposite of 9x^{2}-16, find the opposite of each term.

$$9x^{2}-12x+4-9x^{2}+16\geq 8$$

Combine 9x^{2} and -9x^{2} to get 0.

$$-12x+4+16\geq 8$$

Add 4 and 16 to get 20.

$$-12x+20\geq 8$$

Subtract 20 from both sides.

$$-12x\geq 8-20$$

Subtract 20 from 8 to get -12.

$$-12x\geq -12$$

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

$$x\leq \frac{-12}{-12}$$

Divide -12 by -12 to get 1.

$$x\leq 1$$