Solve {[(1/2x-1)+(1/2x+1)]+(1/2x-1)^2-(1/2x+1)^2}

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

Evaluate

$$-x$$

Solution Steps

Combine \frac{1}{2}x and \frac{1}{2}x to get x.

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

Add -1 and 1 to get 0.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{1}{2}x+1\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{2}x-1\right)^{2}.

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

Combine x and -x to get 0.

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

To find the opposite of \frac{1}{4}x^{2}+x+1, find the opposite of each term.

$$\frac{1}{4}x^{2}+1-\frac{1}{4}x^{2}-x-1$$

Combine \frac{1}{4}x^{2} and -\frac{1}{4}x^{2} to get 0.

$$1-x-1$$

Subtract 1 from 1 to get 0.

$$-x$$

Expand

$$-x$$

Solution Steps

Combine \frac{1}{2}x and \frac{1}{2}x to get x.

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

Add -1 and 1 to get 0.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{1}{2}x+1\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{2}x-1\right)^{2}.

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

Combine x and -x to get 0.

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

To find the opposite of \frac{1}{4}x^{2}+x+1, find the opposite of each term.

$$\frac{1}{4}x^{2}+1-\frac{1}{4}x^{2}-x-1$$

Combine \frac{1}{4}x^{2} and -\frac{1}{4}x^{2} to get 0.

$$1-x-1$$

Subtract 1 from 1 to get 0.

$$-x$$