Resolver y=sqrt{ax^2+4}

$$y = \sqrt { a x ^ { 2 } + 4 }$$

Resolver a

$$a=\frac{y^{2}-4}{x^{2}}$$

$$x\neq 0\text{ and }y\geq 0$$

Pasos para resolver ecuacións radicais

Cambia de lado para que todos os termos variables estean no lado esquerdo.

$$\sqrt{ax^{2}+4}=y$$

Eleva ao cadrado ambos lados da ecuación.

$$x^{2}a+4=y^{2}$$

Resta 4 en ambos lados da ecuación.

$$x^{2}a+4-4=y^{2}-4$$

Se restas 4 a si mesmo, quédache 0.

$$x^{2}a=y^{2}-4$$

Divide ambos lados entre x^{2}.

$$\frac{x^{2}a}{x^{2}}=\frac{y^{2}-4}{x^{2}}$$

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

$$a=\frac{y^{2}-4}{x^{2}}$$

Resolver a (complex solution)

$$\left\{\begin{matrix}a=\frac{y^{2}-4}{x^{2}}\text{, }&x\neq 0\text{ and }\left(y=0\text{ or }arg(y)<\pi \right)\\a\in \mathrm{C}\text{, }&y=2\text{ and }x=0\end{matrix}\right.$$

Resolver x (complex solution)

$$\left\{\begin{matrix}x=-a^{-\frac{1}{2}}\sqrt{y^{2}-4}\text{; }x=a^{-\frac{1}{2}}\sqrt{y^{2}-4}\text{, }&a\neq 0\text{ and }\left(y=0\text{ or }arg(y)<\pi \right)\\x\in \mathrm{C}\text{, }&y=2\text{ and }a=0\end{matrix}\right.$$

Resolver x

$$\left\{\begin{matrix}x=\sqrt{\frac{y^{2}-4}{a}}\text{; }x=-\sqrt{\frac{y^{2}-4}{a}}\text{, }&\left(y\geq 0\text{ and }a<0\text{ and }y<2\right)\text{ or }\left(a>0\text{ and }y>2\right)\\x\in \mathrm{R}\text{, }&y=2\text{ and }a=0\end{matrix}\right.$$