Solve 3/2(x-4)(x+4)-[-x/2(-x+1)+1]geqx^2

$$\frac { 3 } { 2 } ( x - 4 ) ( x + 4 ) - [ - \frac { x } { 2 } ( - x + 1 ) + 1 ] \geq x ^ { 2 }$$

Solve for x

$$x\geq 50$$

Solution Steps

Multiply both sides of the equation by 2. Since 2 is positive, the inequality direction remains the same.

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

Use the distributive property to multiply 3 by x-4.

$$\left(3x-12\right)\left(x+4\right)-2\left(\left(-\frac{x}{2}\right)\left(-x+1\right)+1\right)\geq 2x^{2}$$

Use the distributive property to multiply 3x-12 by x+4 and combine like terms.

$$3x^{2}-48-2\left(\left(-\frac{x}{2}\right)\left(-x+1\right)+1\right)\geq 2x^{2}$$

Express \left(-\frac{x}{2}\right)\left(-x+1\right) as a single fraction.

$$3x^{2}-48-2\left(\frac{-x\left(-x+1\right)}{2}+1\right)\geq 2x^{2}$$

Use the distributive property to multiply -x by -x+1.

$$3x^{2}-48-2\left(\frac{-x\left(-x\right)-x}{2}+1\right)\geq 2x^{2}$$

Multiply -1 and -1 to get 1.

$$3x^{2}-48-2\left(\frac{xx-x}{2}+1\right)\geq 2x^{2}$$

Multiply x and x to get x^{2}.

$$3x^{2}-48-2\left(\frac{x^{2}-x}{2}+1\right)\geq 2x^{2}$$

Divide each term of x^{2}-x by 2 to get \frac{1}{2}x^{2}-\frac{1}{2}x.

$$3x^{2}-48-2\left(\frac{1}{2}x^{2}-\frac{1}{2}x+1\right)\geq 2x^{2}$$

Subtract 2x^{2} from both sides.

$$3x^{2}-48-2\left(\frac{1}{2}x^{2}-\frac{1}{2}x+1\right)-2x^{2}\geq 0$$

Use the distributive property to multiply -2 by \frac{1}{2}x^{2}-\frac{1}{2}x+1.

$$3x^{2}-48-x^{2}+x-2-2x^{2}\geq 0$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$2x^{2}-48+x-2-2x^{2}\geq 0$$

Subtract 2 from -48 to get -50.

$$2x^{2}-50+x-2x^{2}\geq 0$$

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

$$-50+x\geq 0$$

Add 50 to both sides. Anything plus zero gives itself.

$$x\geq 50$$