Solve B=(x+4)^2-x^2-(x-4)^2

$$B = ( x + 4 ) ^ { 2 } - x ^ { 2 } - ( x - 4 ) ^ { 2 }$$

Solve for B

$$B=-x\left(x-16\right)$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$B=x^{2}+8x+16-x^{2}-\left(x-4\right)^{2}$$

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

$$B=8x+16-\left(x-4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$B=8x+16-\left(x^{2}-8x+16\right)$$

To find the opposite of x^{2}-8x+16, find the opposite of each term.

$$B=8x+16-x^{2}+8x-16$$

Combine 8x and 8x to get 16x.

$$B=16x+16-x^{2}-16$$

Subtract 16 from 16 to get 0.

$$B=16x-x^{2}$$

Solve for x (complex solution)

$$x=\sqrt{64-B}+8$$

$$x=-\sqrt{64-B}+8$$

Solve for x

$$x=\sqrt{64-B}+8$$

$$x=-\sqrt{64-B}+8\text{, }B\leq 64$$