Solve -1/8x^2+1/zx+4=0

$$- \frac { 1 } { 8 } x ^ { 2 } + \frac { 1 } { z } x + 4 = 0$$

Solve for z (complex solution)

$$z=-\frac{8x}{32-x^{2}}$$

$$x\neq 0\text{ and }x\neq -4\sqrt{2}\text{ and }x\neq 2^{\frac{5}{2}}$$

Steps for Solving Linear Equation

Variable z cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 8z, the least common multiple of 8,z.

$$-\frac{1}{8}x^{2}\times 8z+8\times 1x+8z\times 4=0$$

Multiply -\frac{1}{8} and 8 to get -1.

$$-x^{2}z+8\times 1x+8z\times 4=0$$

Multiply 8 and 1 to get 8.

$$-x^{2}z+8x+8z\times 4=0$$

Multiply 8 and 4 to get 32.

$$-x^{2}z+8x+32z=0$$

Subtract 8x from both sides. Anything subtracted from zero gives its negation.

$$-x^{2}z+32z=-8x$$

Combine all terms containing z.

$$\left(-x^{2}+32\right)z=-8x$$

The equation is in standard form.

$$\left(32-x^{2}\right)z=-8x$$

Divide both sides by -x^{2}+32.

$$\frac{\left(32-x^{2}\right)z}{32-x^{2}}=-\frac{8x}{32-x^{2}}$$

Dividing by -x^{2}+32 undoes the multiplication by -x^{2}+32.

$$z=-\frac{8x}{32-x^{2}}$$

Variable z cannot be equal to 0.

$$z=-\frac{8x}{32-x^{2}}\text{, }z\neq 0$$

Solve for z

$$z=-\frac{8x}{32-x^{2}}$$

$$x\neq 0\text{ and }|x|\neq 4\sqrt{2}$$

Steps for Solving Linear Equation

Variable z cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 8z, the least common multiple of 8,z.

$$-\frac{1}{8}x^{2}\times 8z+8\times 1x+8z\times 4=0$$

Multiply -\frac{1}{8} and 8 to get -1.

$$-x^{2}z+8\times 1x+8z\times 4=0$$

Multiply 8 and 1 to get 8.

$$-x^{2}z+8x+8z\times 4=0$$

Multiply 8 and 4 to get 32.

$$-x^{2}z+8x+32z=0$$

Subtract 8x from both sides. Anything subtracted from zero gives its negation.

$$-x^{2}z+32z=-8x$$

Combine all terms containing z.

$$\left(-x^{2}+32\right)z=-8x$$

The equation is in standard form.

$$\left(32-x^{2}\right)z=-8x$$

Divide both sides by 32-x^{2}.

$$\frac{\left(32-x^{2}\right)z}{32-x^{2}}=-\frac{8x}{32-x^{2}}$$

Dividing by 32-x^{2} undoes the multiplication by 32-x^{2}.

$$z=-\frac{8x}{32-x^{2}}$$

Variable z cannot be equal to 0.

$$z=-\frac{8x}{32-x^{2}}\text{, }z\neq 0$$

Solve for x

$$x=-\frac{4\left(\sqrt{2z^{2}+1}-1\right)}{z}$$

$$x=\frac{4\left(\sqrt{2z^{2}+1}+1\right)}{z}\text{, }z\neq 0$$