Resolver (x^2-3)sqrt{2x+y}=0

$$( { x }^{ 2 } -3) \sqrt{ 2x+y } =0$$

Resolver y

$$\left\{\begin{matrix}\\y=-2x\text{, }&\text{unconditionally}\\y\geq -2x\text{, }&|x|=\sqrt{3}\end{matrix}\right.$$

Pasos para resolver ecuacións radicais

Divide ambos lados entre x^{2}-3.

$$\frac{\left(x^{2}-3\right)\sqrt{y+2x}}{x^{2}-3}=\frac{0}{x^{2}-3}$$

A división entre x^{2}-3 desfai a multiplicación por x^{2}-3.

$$\sqrt{y+2x}=\frac{0}{x^{2}-3}$$

Divide 0 entre x^{2}-3.

$$\sqrt{y+2x}=0$$

Eleva ao cadrado ambos lados da ecuación.

$$y+2x=0$$

Resta 2x en ambos lados da ecuación.

$$y+2x-2x=-2x$$

Se restas 2x a si mesmo, quédache 0.

$$y=-2x$$

Resolver x (complex solution)

$$x=-\sqrt{3}\approx -1.732050808$$

$$x=\sqrt{3}\approx 1.732050808$$

$$x=-\frac{y}{2}$$

Resolver y (complex solution)

$$\left\{\begin{matrix}\\y=-2x\text{, }&\text{unconditionally}\\y\in \mathrm{C}\text{, }&x=-\sqrt{3}\text{ or }x=\sqrt{3}\end{matrix}\right.$$

Resolver x

$$\left\{\begin{matrix}\\x=-\frac{y}{2}\text{, }&\text{unconditionally}\\x=-\sqrt{3}\text{, }&y\geq 2\sqrt{3}\\x=\sqrt{3}\approx 1.732050808\text{, }&y\geq -2\sqrt{3}\end{matrix}\right.$$