Solve (x+3)(x-5)-(x-3)^2geq0

$$( x + 3 ) ( x - 5 ) - ( x - 3 ) ^ { 2 } \geq 0$$

Solve for x

$$x\geq 6$$

Solution Steps

Use the distributive property to multiply x+3 by x-5 and combine like terms.

$$x^{2}-2x-15-\left(x-3\right)^{2}\geq 0$$

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

$$x^{2}-2x-15-\left(x^{2}-6x+9\right)\geq 0$$

To find the opposite of x^{2}-6x+9, find the opposite of each term.

$$x^{2}-2x-15-x^{2}+6x-9\geq 0$$

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

$$-2x-15+6x-9\geq 0$$

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

$$4x-15-9\geq 0$$

Subtract 9 from -15 to get -24.

$$4x-24\geq 0$$

Add 24 to both sides. Anything plus zero gives itself.

$$4x\geq 24$$

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

$$x\geq \frac{24}{4}$$

Divide 24 by 4 to get 6.

$$x\geq 6$$