Solve B=(3x-1/2)^2-(3x+1/2)^2

$$B = ( 3 x - \frac { 1 } { 2 } ) ^ { 2 } - ( 3 x + \frac { 1 } { 2 } ) ^ { 2 }$$

Solve for B

$$B=-6x$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-\frac{1}{2}\right)^{2}.

$$B=9x^{2}-3x+\frac{1}{4}-\left(3x+\frac{1}{2}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+\frac{1}{2}\right)^{2}.

$$B=9x^{2}-3x+\frac{1}{4}-\left(9x^{2}+3x+\frac{1}{4}\right)$$

To find the opposite of 9x^{2}+3x+\frac{1}{4}, find the opposite of each term.

$$B=9x^{2}-3x+\frac{1}{4}-9x^{2}-3x-\frac{1}{4}$$

Combine 9x^{2} and -9x^{2} to get 0.

$$B=-3x+\frac{1}{4}-3x-\frac{1}{4}$$

Combine -3x and -3x to get -6x.

$$B=-6x+\frac{1}{4}-\frac{1}{4}$$

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

$$B=-6x$$

Solve for x

$$x=-\frac{B}{6}$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-\frac{1}{2}\right)^{2}.

$$B=9x^{2}-3x+\frac{1}{4}-\left(3x+\frac{1}{2}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+\frac{1}{2}\right)^{2}.

$$B=9x^{2}-3x+\frac{1}{4}-\left(9x^{2}+3x+\frac{1}{4}\right)$$

To find the opposite of 9x^{2}+3x+\frac{1}{4}, find the opposite of each term.

$$B=9x^{2}-3x+\frac{1}{4}-9x^{2}-3x-\frac{1}{4}$$

Combine 9x^{2} and -9x^{2} to get 0.

$$B=-3x+\frac{1}{4}-3x-\frac{1}{4}$$

Combine -3x and -3x to get -6x.

$$B=-6x+\frac{1}{4}-\frac{1}{4}$$

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

$$B=-6x$$

Swap sides so that all variable terms are on the left hand side.

$$-6x=B$$

Divide both sides by -6.

$$\frac{-6x}{-6}=\frac{B}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x=\frac{B}{-6}$$

Divide B by -6.

$$x=-\frac{B}{6}$$