Solve sqrt{d+rsqrt{q}}=sqrt{r}+sqrt{r}

$$\sqrt { d + r \sqrt { q } } = \sqrt { r } + \sqrt { r }$$

Solve for d

$$d=-\left(\sqrt{q}-4\right)r$$

$$r\geq 0\text{ and }q>0$$

Steps for Solving Radical Equations

Square both sides of the equation.

$$d+\sqrt{q}r=4r$$

Subtract r\sqrt{q} from both sides of the equation.

$$d+\sqrt{q}r-\sqrt{q}r=4r-\sqrt{q}r$$

Subtracting r\sqrt{q} from itself leaves 0.

$$d=4r-\sqrt{q}r$$

Subtract r\sqrt{q} from 4r.

$$d=\left(-\sqrt{q}+4\right)r$$

Solve for d (complex solution)

$$d=-\left(\sqrt{q}-4\right)r$$

Solve for q (complex solution)

$$\left\{\begin{matrix}q=\left(\frac{d}{r}-4\right)^{2}\text{, }&\left(d=4r\text{ or }arg(\frac{d}{r}-4)\geq \pi \right)\text{ and }r\neq 0\\q\in \mathrm{C}\text{, }&d=0\text{ and }r=0\end{matrix}\right.$$

Solve for q

$$\left\{\begin{matrix}q=\left(\frac{d}{r}-4\right)^{2}\text{, }&\left(d>0\text{ and }d\leq 4r\right)\text{ or }\left(r>0\text{ and }d\leq 2r\right)\\q\geq 0\text{, }&r=0\text{ and }d=0\end{matrix}\right.$$