Solve (x^{2}+2y)^2=(x^4+2x^2y+m+4y^2)

$$( x ^ { 2 } + 2 y ) ^ { 2 } = ( x ^ { 4 } + 2 x ^ { 2 } y + m + 4 y ^ { 2 } )$$

Solve for m

$$m=2yx^{2}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{2}+2y\right)^{2}.

$$\left(x^{2}\right)^{2}+4x^{2}y+4y^{2}=x^{4}+2x^{2}y+m+4y^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract x^{4} from both sides.

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

Combine x^{4} and -x^{4} to get 0.

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

Subtract 2x^{2}y from both sides.

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

Combine 4x^{2}y and -2x^{2}y to get 2x^{2}y.

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

Subtract 4y^{2} from both sides.

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

Combine 4y^{2} and -4y^{2} to get 0.

$$m=2x^{2}y$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=-\frac{y^{-\frac{1}{2}}\sqrt{2m}}{2}\text{; }x=\frac{y^{-\frac{1}{2}}\sqrt{2m}}{2}\text{, }&y\neq 0\\x\in \mathrm{C}\text{, }&m=0\text{ and }y=0\end{matrix}\right.$$

Solve for x

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