Lutasin ang sqrt{x^4}sqrt{y}=x

$$\sqrt { x ^ { 4 } } \sqrt { y } = x$$

I-solve ang y

$$\left\{\begin{matrix}y=\frac{1}{x^{2}}\text{, }&x>0\\y\geq 0\text{, }&x=0\end{matrix}\right.$$

Mga Hakbang sa Pagso-solve ng Mga Radical Equation

I-divide ang magkabilang dulo ng equation gamit ang \sqrt{x^{4}}.

$$\frac{\sqrt{x^{4}}\sqrt{y}}{\sqrt{x^{4}}}=\frac{x}{\sqrt{x^{4}}}$$

Kapag na-divide gamit ang \sqrt{x^{4}}, ma-a-undo ang multiplication gamit ang \sqrt{x^{4}}.

$$\sqrt{y}=\frac{x}{\sqrt{x^{4}}}$$

I-divide ang x gamit ang \sqrt{x^{4}}.

$$\sqrt{y}=\frac{1}{x}$$

I-square ang magkabilang dulo ng equation.

$$y=\frac{1}{x^{2}}$$

I-solve ang x (complex solution)

$$\left\{\begin{matrix}\\x=0\text{, }&\text{unconditionally}\\x=y^{-\frac{1}{2}}\text{, }&|-arg(y^{-\frac{1}{2}})+arg(\sqrt{\frac{1}{y^{2}}}\sqrt{y})|<\pi \text{ and }y\neq 0\\x=-y^{-\frac{1}{2}}\text{, }&|-arg(-y^{-\frac{1}{2}})+arg(\sqrt{\frac{1}{y^{2}}}\sqrt{y})|<\pi \text{ and }y\neq 0\end{matrix}\right.$$

I-solve ang y (complex solution)

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

I-solve ang x

$$\left\{\begin{matrix}x=0\text{, }&y\geq 0\\x=\frac{1}{\sqrt{y}}\text{, }&y>0\end{matrix}\right.$$