Solve (1/2-v)^2-(1/2-v)*(1/2+v)+(v+1/2)^2

$$( \frac { 1 } { 2 } - v ) ^ { 2 } - ( \frac { 1 } { 2 } - v ) \cdot ( \frac { 1 } { 2 } + v ) + ( v + \frac { 1 } { 2 } ) ^ { 2 }$$

Evaluate

$$3v^{2}+\frac{1}{4}$$

Solution Steps

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

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

Consider \left(\frac{1}{2}-v\right)\left(\frac{1}{2}+v\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 \frac{1}{2}.

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

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

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

Subtract \frac{1}{4} from \frac{1}{4} to get 0.

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

Combine v^{2} and v^{2} to get 2v^{2}.

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

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

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

Combine 2v^{2} and v^{2} to get 3v^{2}.

$$-v+3v^{2}+v+\frac{1}{4}$$

Combine -v and v to get 0.

$$3v^{2}+\frac{1}{4}$$

Expand

$$3v^{2}+\frac{1}{4}$$

Solution Steps

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

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

Consider \left(\frac{1}{2}-v\right)\left(\frac{1}{2}+v\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 \frac{1}{2}.

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

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

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

Subtract \frac{1}{4} from \frac{1}{4} to get 0.

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

Combine v^{2} and v^{2} to get 2v^{2}.

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

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

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

Combine 2v^{2} and v^{2} to get 3v^{2}.

$$-v+3v^{2}+v+\frac{1}{4}$$

Combine -v and v to get 0.

$$3v^{2}+\frac{1}{4}$$