Solve (3/2-x/2)^2-x/2(x/2-3)+(x-frac{3}{2})(x+frac{3}{2})-frac{3}{2}v^2

$$( \frac { 3 } { 2 } - \frac { x } { 2 } ) ^ { 2 } - \frac { x } { 2 } ( \frac { x } { 2 } - 3 ) + ( x - \frac { 3 } { 2 } ) ( x + \frac { 3 } { 2 } ) - \frac { 3 } { 2 } v ^ { 2 }$$

Evaluate

$$-\frac{3v^{2}}{2}+x^{2}$$

Short Solution Steps

Since \frac{3}{2} and \frac{x}{2} have the same denominator, add them by adding their numerators.

$$\left(\frac{3+x}{2}\right)^{2}-\frac{x}{2}\left(\frac{x}{2}-3\right)+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

To raise \frac{3+x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3+x\right)^{2}}{2^{2}}-\frac{x}{2}\left(\frac{x}{2}-3\right)+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{2}{2}.

$$\frac{\left(3+x\right)^{2}}{2^{2}}-\frac{x}{2}\left(\frac{x}{2}-\frac{3\times 2}{2}\right)+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

Since \frac{x}{2} and \frac{3\times 2}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(3+x\right)^{2}}{2^{2}}-\frac{x}{2}\times \frac{x-3\times 2}{2}+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

Do the multiplications in x-3\times 2.

$$\frac{\left(3+x\right)^{2}}{2^{2}}-\frac{x}{2}\times \frac{x-6}{2}+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

Multiply \frac{x}{2} times \frac{x-6}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(3+x\right)^{2}}{2^{2}}-\frac{x\left(x-6\right)}{2\times 2}+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

Multiply 2 and 2 to get 4.

$$\frac{\left(3+x\right)^{2}}{2^{2}}-\frac{x\left(x-6\right)}{4}+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Expand 2^{2}.

$$\frac{\left(3+x\right)^{2}}{4}-\frac{x\left(x-6\right)}{4}+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

Since \frac{\left(3+x\right)^{2}}{4} and \frac{x\left(x-6\right)}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(3+x\right)^{2}-x\left(x-6\right)}{4}+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

Do the multiplications in \left(3+x\right)^{2}-x\left(x-6\right).

$$\frac{9+6x+x^{2}-x^{2}+6x}{4}+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

Combine like terms in 9+6x+x^{2}-x^{2}+6x.

$$\frac{9+12x}{4}+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

Consider \left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\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{3}{2}.

$$\frac{9+12x}{4}+x^{2}-\frac{9}{4}-\frac{3}{2}v^{2}$$

Since \frac{9+12x}{4} and \frac{9}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{9+12x-9}{4}+x^{2}-\frac{3}{2}v^{2}$$

Combine like terms in 9+12x-9.

$$\frac{12x}{4}+x^{2}-\frac{3}{2}v^{2}$$

Divide 12x by 4 to get 3x.

$$3x+x^{2}-\frac{3}{2}v^{2}$$

Expand

$$-\frac{3v^{2}}{2}+x^{2}$$

Short Solution Steps

Since \frac{3}{2} and \frac{x}{2} have the same denominator, add them by adding their numerators.

$$\left(\frac{3+x}{2}\right)^{2}-\frac{x}{2}\left(\frac{x}{2}-3\right)+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

To raise \frac{3+x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3+x\right)^{2}}{2^{2}}-\frac{x}{2}\left(\frac{x}{2}-3\right)+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{2}{2}.

$$\frac{\left(3+x\right)^{2}}{2^{2}}-\frac{x}{2}\left(\frac{x}{2}-\frac{3\times 2}{2}\right)+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

Since \frac{x}{2} and \frac{3\times 2}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(3+x\right)^{2}}{2^{2}}-\frac{x}{2}\times \frac{x-3\times 2}{2}+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

Do the multiplications in x-3\times 2.

$$\frac{\left(3+x\right)^{2}}{2^{2}}-\frac{x}{2}\times \frac{x-6}{2}+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

Multiply \frac{x}{2} times \frac{x-6}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(3+x\right)^{2}}{2^{2}}-\frac{x\left(x-6\right)}{2\times 2}+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

Multiply 2 and 2 to get 4.

$$\frac{\left(3+x\right)^{2}}{2^{2}}-\frac{x\left(x-6\right)}{4}+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Expand 2^{2}.

$$\frac{\left(3+x\right)^{2}}{4}-\frac{x\left(x-6\right)}{4}+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

Since \frac{\left(3+x\right)^{2}}{4} and \frac{x\left(x-6\right)}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(3+x\right)^{2}-x\left(x-6\right)}{4}+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

Do the multiplications in \left(3+x\right)^{2}-x\left(x-6\right).

$$\frac{9+6x+x^{2}-x^{2}+6x}{4}+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

Combine like terms in 9+6x+x^{2}-x^{2}+6x.

$$\frac{9+12x}{4}+\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)-\frac{3}{2}v^{2}$$

Consider \left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\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{3}{2}.

$$\frac{9+12x}{4}+x^{2}-\frac{9}{4}-\frac{3}{2}v^{2}$$

Since \frac{9+12x}{4} and \frac{9}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{9+12x-9}{4}+x^{2}-\frac{3}{2}v^{2}$$

Combine like terms in 9+12x-9.

$$\frac{12x}{4}+x^{2}-\frac{3}{2}v^{2}$$

Divide 12x by 4 to get 3x.

$$3x+x^{2}-\frac{3}{2}v^{2}$$