Solve x^2operatorname{dy}4ydx=0

$$x ^ { 2 } \operatorname { dy } 4 y d x = 0$$

Solve for d (complex solution)

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{C}\text{, }&x=0\text{ or }y=0\text{ or }y_{4}=0\end{matrix}\right.$$

Steps by Finding Square Root

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$x^{3}dy_{4}yd=0$$

Multiply d and d to get d^{2}.

$$x^{3}d^{2}y_{4}y=0$$

Dividing by x^{3}y_{4}y undoes the multiplication by x^{3}y_{4}y.

$$d^{2}=\frac{0}{yy_{4}x^{3}}$$

Divide 0 by x^{3}y_{4}y.

$$d^{2}=0$$

Take the square root of both sides of the equation.

$$d=0 d=0$$

The equation is now solved. Solutions are the same.

$$d=0$$

Steps Using the Quadratic Formula

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$x^{3}dy_{4}yd=0$$

Multiply d and d to get d^{2}.

$$x^{3}d^{2}y_{4}y=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$yy_{4}x^{3}d^{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute x^{3}y_{4}y for a, 0 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$d=\frac{0±\sqrt{0^{2}}}{2yy_{4}x^{3}}$$

Take the square root of 0^{2}.

$$d=\frac{0±0}{2yy_{4}x^{3}}$$

Multiply 2 times x^{3}y_{4}y.

$$d=\frac{0}{2yy_{4}x^{3}}$$

Divide 0 by 2x^{3}y_{4}y.

$$d=0$$

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=0\text{, }&\text{unconditionally}\\x\in \mathrm{C}\text{, }&d=0\text{ or }y=0\text{ or }y_{4}=0\end{matrix}\right.$$

Solve for d

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&x=0\text{ or }y=0\text{ or }y_{4}=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}\\x=0\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&d=0\text{ or }y=0\text{ or }y_{4}=0\end{matrix}\right.$$