Solve sqrt{x+1}=sqrt[3]{y/2}-4/2

$$\sqrt { x + 1 } = \sqrt[ 3 ] { \frac { y } { 2 } } - \frac { 4 } { 2 }$$

Solve for x

$$x=\frac{\sqrt[3]{2}y^{\frac{2}{3}}-4\times 2^{\frac{2}{3}}\sqrt[3]{y}+6}{2}$$

$$\frac{2^{\frac{2}{3}}\sqrt[3]{y}}{2}-2\geq 0$$

Steps for Solving Radical Equations

Square both sides of the equation.

$$x+1=\frac{\left(\sqrt[3]{y}-2\sqrt[3]{2}\right)^{2}}{2^{\frac{2}{3}}}$$

Subtract 1 from both sides of the equation.

$$x+1-1=\frac{\left(\sqrt[3]{y}-2\sqrt[3]{2}\right)^{2}}{2^{\frac{2}{3}}}-1$$

Subtracting 1 from itself leaves 0.

$$x=\frac{\left(\sqrt[3]{y}-2\sqrt[3]{2}\right)^{2}}{2^{\frac{2}{3}}}-1$$

Subtract 1 from \frac{\left(\sqrt[3]{y}-2\sqrt[3]{2}\right)^{2}}{2^{\frac{2}{3}}}.

$$x=\frac{y^{\frac{2}{3}}}{2^{\frac{2}{3}}}-\frac{4\sqrt[3]{y}}{\sqrt[3]{2}}+3$$

Solve for x (complex solution)

$$x=\frac{\sqrt[3]{2}y^{\frac{2}{3}}-4\times 2^{\frac{2}{3}}\sqrt[3]{y}+6}{2}$$

$$arg(-\frac{2^{\frac{2}{3}}\sqrt[3]{y}}{2}+2)\geq \pi \text{ or }y=16$$

Solve for y (complex solution)

$$y=2\left(\sqrt{x+1}+2\right)^{3}$$

$$arg(\sqrt{x+1}+2)<\frac{2\pi }{3}$$

Solve for y

$$y=2\left(\sqrt{x+1}+2\right)^{3}$$

$$x\geq -1$$