Solve (2-x)^2-7x+2geq(x+4)(x-4)

$$( 2 - x ) ^ { 2 } - 7 x + 2 \geq ( x + 4 ) ( x - 4 )$$

Solve for x

$$x\leq 2$$

Solution Steps

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

$$4-4x+x^{2}-7x+2\geq \left(x+4\right)\left(x-4\right)$$

Combine -4x and -7x to get -11x.

$$4-11x+x^{2}+2\geq \left(x+4\right)\left(x-4\right)$$

Add 4 and 2 to get 6.

$$6-11x+x^{2}\geq \left(x+4\right)\left(x-4\right)$$

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

$$6-11x+x^{2}\geq x^{2}-16$$

Subtract x^{2} from both sides.

$$6-11x+x^{2}-x^{2}\geq -16$$

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

$$6-11x\geq -16$$

Subtract 6 from both sides.

$$-11x\geq -16-6$$

Subtract 6 from -16 to get -22.

$$-11x\geq -22$$

Divide both sides by -11. Since -11 is negative, the inequality direction is changed.

$$x\leq \frac{-22}{-11}$$

Divide -22 by -11 to get 2.

$$x\leq 2$$