Solve x^2-3ax+8/x-2=0

$$\frac{ { x }^{ 2 } -3ax+8 }{ x-2 } =0$$

Solve for a

$$a=\frac{x}{3}+\frac{8}{3x}$$

$$x\neq 0\text{ and }x\neq 2$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x-2.

$$x^{2}-3ax+8=0$$

Subtract x^{2} from both sides. Anything subtracted from zero gives its negation.

$$-3ax+8=-x^{2}$$

Subtract 8 from both sides.

$$-3ax=-x^{2}-8$$

The equation is in standard form.

$$\left(-3x\right)a=-x^{2}-8$$

Divide both sides by -3x.

$$\frac{\left(-3x\right)a}{-3x}=\frac{-x^{2}-8}{-3x}$$

Dividing by -3x undoes the multiplication by -3x.

$$a=\frac{-x^{2}-8}{-3x}$$

Divide -x^{2}-8 by -3x.

$$a=\frac{x}{3}+\frac{8}{3x}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=\frac{\sqrt{9a^{2}-32}+3a}{2}\text{, }&\text{unconditionally}\\x=\frac{-\sqrt{9a^{2}-32}+3a}{2}\text{, }&a\neq 2\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{-\sqrt{9a^{2}-32}+3a}{2}\text{, }&a\leq -\frac{4\sqrt{2}}{3}\text{ or }\left(a\neq 2\text{ and }a\geq \frac{4\sqrt{2}}{3}\right)\\x=\frac{\sqrt{9a^{2}-32}+3a}{2}\text{, }&|a|\geq \frac{4\sqrt{2}}{3}\end{matrix}\right.$$