Solve (x-3)^2geq(x-7)(x+5)

$$( x - 3 ) ^ { 2 } \geq ( x - 7 ) ( x + 5 )$$

Solve for x

$$x\leq 11$$

Solution Steps

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

$$x^{2}-6x+9\geq \left(x-7\right)\left(x+5\right)$$

Use the distributive property to multiply x-7 by x+5 and combine like terms.

$$x^{2}-6x+9\geq x^{2}-2x-35$$

Subtract x^{2} from both sides.

$$x^{2}-6x+9-x^{2}\geq -2x-35$$

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

$$-6x+9\geq -2x-35$$

Add 2x to both sides.

$$-6x+9+2x\geq -35$$

Combine -6x and 2x to get -4x.

$$-4x+9\geq -35$$

Subtract 9 from both sides.

$$-4x\geq -35-9$$

Subtract 9 from -35 to get -44.

$$-4x\geq -44$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x\leq \frac{-44}{-4}$$

Divide -44 by -4 to get 11.

$$x\leq 11$$