Solve a+bsqrt{3}=(m+nsqrt{3})^2

$$a + b \sqrt { 3 } = ( m + n \sqrt { 3 } ) ^ { 2 }$$

Solve for b

$$b=-\frac{\sqrt{3}\left(-\left(\sqrt{3}n+m\right)^{2}+a\right)}{3}$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(m+n\sqrt{3}\right)^{2}.

$$a+b\sqrt{3}=m^{2}+2mn\sqrt{3}+n^{2}\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$a+b\sqrt{3}=m^{2}+2mn\sqrt{3}+n^{2}\times 3$$

Subtract a from both sides.

$$b\sqrt{3}=m^{2}+2mn\sqrt{3}+n^{2}\times 3-a$$

The equation is in standard form.

$$\sqrt{3}b=2\sqrt{3}mn+3n^{2}+m^{2}-a$$

Divide both sides by \sqrt{3}.

$$\frac{\sqrt{3}b}{\sqrt{3}}=\frac{2\sqrt{3}mn+3n^{2}+m^{2}-a}{\sqrt{3}}$$

Dividing by \sqrt{3} undoes the multiplication by \sqrt{3}.

$$b=\frac{2\sqrt{3}mn+3n^{2}+m^{2}-a}{\sqrt{3}}$$

Divide m^{2}+2mn\sqrt{3}+3n^{2}-a by \sqrt{3}.

$$b=\frac{\sqrt{3}\left(2\sqrt{3}mn+3n^{2}+m^{2}-a\right)}{3}$$

Solve for a

$$a=\left(\sqrt{3}n+m\right)^{2}-\sqrt{3}b$$