Solve 1ax^2-4y^2-8x+12=0

$$1 a x ^ { 2 } - 4 y ^ { 2 } - 8 x + 12 = 0$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=-\frac{4\left(3-y^{2}-2x\right)}{x^{2}}\text{, }&x\neq 0\\a\in \mathrm{C}\text{, }&\left(y=\sqrt{3}\text{ or }y=-\sqrt{3}\right)\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Add 4y^{2} to both sides. Anything plus zero gives itself.

$$1ax^{2}-8x+12=4y^{2}$$

Add 8x to both sides.

$$1ax^{2}+12=4y^{2}+8x$$

Subtract 12 from both sides.

$$1ax^{2}=4y^{2}+8x-12$$

Reorder the terms.

$$ax^{2}=8x+4y^{2}-12$$

The equation is in standard form.

$$x^{2}a=8x+4y^{2}-12$$

Divide both sides by x^{2}.

$$\frac{x^{2}a}{x^{2}}=\frac{8x+4y^{2}-12}{x^{2}}$$

Dividing by x^{2} undoes the multiplication by x^{2}.

$$a=\frac{8x+4y^{2}-12}{x^{2}}$$

Divide 8x+4y^{2}-12 by x^{2}.

$$a=\frac{4\left(2x+y^{2}-3\right)}{x^{2}}$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{4\left(3-y^{2}-2x\right)}{x^{2}}\text{, }&x\neq 0\\a\in \mathrm{R}\text{, }&x=0\text{ and }|y|=\sqrt{3}\end{matrix}\right.$$

Steps for Solving Linear Equation

Add 4y^{2} to both sides. Anything plus zero gives itself.

$$1ax^{2}-8x+12=4y^{2}$$

Add 8x to both sides.

$$1ax^{2}+12=4y^{2}+8x$$

Subtract 12 from both sides.

$$1ax^{2}=4y^{2}+8x-12$$

Reorder the terms.

$$ax^{2}=8x+4y^{2}-12$$

The equation is in standard form.

$$x^{2}a=8x+4y^{2}-12$$

Divide both sides by x^{2}.

$$\frac{x^{2}a}{x^{2}}=\frac{8x+4y^{2}-12}{x^{2}}$$

Dividing by x^{2} undoes the multiplication by x^{2}.

$$a=\frac{8x+4y^{2}-12}{x^{2}}$$

Divide 8x+4y^{2}-12 by x^{2}.

$$a=\frac{4\left(2x+y^{2}-3\right)}{x^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{2\left(\sqrt{ay^{2}-3a+4}+2\right)}{a}\text{; }x=\frac{2\left(-\sqrt{ay^{2}-3a+4}+2\right)}{a}\text{, }&a\neq 0\\x=\frac{3-y^{2}}{2}\text{, }&a=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{2\left(\sqrt{ay^{2}-3a+4}+2\right)}{a}\text{; }x=\frac{2\left(-\sqrt{ay^{2}-3a+4}+2\right)}{a}\text{, }&\left(a=-\frac{4}{y^{2}-3}\text{ and }y<-\sqrt{3}\right)\text{ or }\left(|y|\leq \sqrt{3}\text{ and }a\neq 0\text{ and }a\leq -\frac{4}{y^{2}-3}\right)\text{ or }\left(a=-\frac{4}{y^{2}-3}\text{ and }|y|\neq \sqrt{3}\right)\text{ or }\left(a\neq 0\text{ and }a\geq -\frac{4}{y^{2}-3}\text{ and }|y|\geq \sqrt{3}\right)\text{ or }\left(a\neq 0\text{ and }|y|=\sqrt{3}\right)\\x=\frac{3-y^{2}}{2}\text{, }&a=0\end{matrix}\right.$$