Solve 4m/2h=frac{sqrt{p-k}}{m}

$$\frac { 4 m } { 2 h } = \frac { \sqrt { p - k } } { m }$$

Solve for h (complex solution)

$$h=2\left(p-k\right)^{-\frac{1}{2}}m^{2}$$

$$m\neq 0\text{ and }p\neq k$$

Steps for Solving Linear Equation

Variable h cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2hm, the least common multiple of 2h,m.

$$m\times 4m=2h\sqrt{p-k}$$

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

$$m^{2}\times 4=2h\sqrt{p-k}$$

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

$$2h\sqrt{p-k}=m^{2}\times 4$$

The equation is in standard form.

$$2\sqrt{p-k}h=4m^{2}$$

Divide both sides by 2\sqrt{p-k}.

$$\frac{2\sqrt{p-k}h}{2\sqrt{p-k}}=\frac{4m^{2}}{2\sqrt{p-k}}$$

Dividing by 2\sqrt{p-k} undoes the multiplication by 2\sqrt{p-k}.

$$h=\frac{4m^{2}}{2\sqrt{p-k}}$$

Divide 4m^{2} by 2\sqrt{p-k}.

$$h=2\left(p-k\right)^{-\frac{1}{2}}m^{2}$$

Variable h cannot be equal to 0.

$$h=2\left(p-k\right)^{-\frac{1}{2}}m^{2}\text{, }h\neq 0$$

Solve for h

$$h=\frac{2m^{2}}{\sqrt{p-k}}$$

$$m\neq 0\text{ and }p>k$$

Steps for Solving Linear Equation

Variable h cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2hm, the least common multiple of 2h,m.

$$m\times 4m=2h\sqrt{p-k}$$

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

$$m^{2}\times 4=2h\sqrt{p-k}$$

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

$$2h\sqrt{p-k}=m^{2}\times 4$$

The equation is in standard form.

$$2\sqrt{p-k}h=4m^{2}$$

Divide both sides by 2\sqrt{p-k}.

$$\frac{2\sqrt{p-k}h}{2\sqrt{p-k}}=\frac{4m^{2}}{2\sqrt{p-k}}$$

Dividing by 2\sqrt{p-k} undoes the multiplication by 2\sqrt{p-k}.

$$h=\frac{4m^{2}}{2\sqrt{p-k}}$$

Divide 4m^{2} by 2\sqrt{p-k}.

$$h=\frac{2m^{2}}{\sqrt{p-k}}$$

Variable h cannot be equal to 0.

$$h=\frac{2m^{2}}{\sqrt{p-k}}\text{, }h\neq 0$$

Solve for k

$$k=-\frac{4m^{4}}{h^{2}}+p$$

$$h>0\text{ and }m\neq 0$$

Steps for Solving Radical Equations

Multiply both sides of the equation by 2hm, the least common multiple of 2h,m.

$$m\times 4m=2h\sqrt{p-k}$$

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

$$m^{2}\times 4=2h\sqrt{p-k}$$

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

$$2h\sqrt{p-k}=m^{2}\times 4$$

Divide both sides by 2h.

$$\frac{2h\sqrt{-k+p}}{2h}=\frac{4m^{2}}{2h}$$

Dividing by 2h undoes the multiplication by 2h.

$$\sqrt{-k+p}=\frac{4m^{2}}{2h}$$

Divide 4m^{2} by 2h.

$$\sqrt{-k+p}=\frac{2m^{2}}{h}$$

Square both sides of the equation.

$$-k+p=\frac{4m^{4}}{h^{2}}$$

Subtract p from both sides of the equation.

$$-k+p-p=\frac{4m^{4}}{h^{2}}-p$$

Subtracting p from itself leaves 0.

$$-k=\frac{4m^{4}}{h^{2}}-p$$

Divide both sides by -1.

$$\frac{-k}{-1}=\frac{\frac{4m^{4}}{h^{2}}-p}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$k=\frac{\frac{4m^{4}}{h^{2}}-p}{-1}$$

Divide -p+\frac{4m^{4}}{h^{2}} by -1.

$$k=-\frac{4m^{4}}{h^{2}}+p$$

Solve for k (complex solution)

$$k=-\frac{4m^{4}}{h^{2}}+p$$

$$|arg(h\sqrt{\frac{m^{4}}{h^{2}}})-arg(m^{2})|<\pi \text{ and }h\neq 0\text{ and }m\neq 0$$