Solve (x-m)^2=x^2+x+a

$$( x - m ) ^ { 2 } = x ^ { 2 } + x + a$$

Solve for a

$$a=-\left(m\left(2x-m\right)+x\right)$$

Solution Steps

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

$$x^{2}-2xm+m^{2}=x^{2}+x+a$$

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

$$x^{2}+x+a=x^{2}-2xm+m^{2}$$

Subtract x^{2} from both sides.

$$x+a=x^{2}-2xm+m^{2}-x^{2}$$

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

$$x+a=-2xm+m^{2}$$

Subtract x from both sides.

$$a=-2xm+m^{2}-x$$

Solve for m (complex solution)

$$m=\sqrt{x^{2}+x+a}+x$$

$$m=-\sqrt{x^{2}+x+a}+x$$

Solve for m

$$m=\sqrt{x^{2}+x+a}+x$$

$$m=-\sqrt{x^{2}+x+a}+x\text{, }a\geq -\left(x^{2}+x\right)$$