Solve (a-R)^2+3a^2=R^2

$$( a - R ) ^ { 2 } + 3 a ^ { 2 } = R ^ { 2 }$$

Solve for R

$$\left\{\begin{matrix}\\R=2a\text{, }&\text{unconditionally}\\R\in \mathrm{R}\text{, }&a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-R\right)^{2}.

$$a^{2}-2aR+R^{2}+3a^{2}=R^{2}$$

Combine a^{2} and 3a^{2} to get 4a^{2}.

$$4a^{2}-2aR+R^{2}=R^{2}$$

Subtract R^{2} from both sides.

$$4a^{2}-2aR+R^{2}-R^{2}=0$$

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

$$4a^{2}-2aR=0$$

Subtract 4a^{2} from both sides. Anything subtracted from zero gives its negation.

$$-2aR=-4a^{2}$$

The equation is in standard form.

$$\left(-2a\right)R=-4a^{2}$$

Divide both sides by -2a.

$$\frac{\left(-2a\right)R}{-2a}=-\frac{4a^{2}}{-2a}$$

Dividing by -2a undoes the multiplication by -2a.

$$R=-\frac{4a^{2}}{-2a}$$

Divide -4a^{2} by -2a.

$$R=2a$$

Solve for a

$$a=\frac{R}{2}$$

$$a=0$$