Solve frac{sqrt{2}}{8}=frac{sqrt{x-a}}{bx^2}

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

Solve for b (complex solution)

$$b=\frac{4\sqrt{2\left(x-a\right)}}{x^{2}}$$

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

Steps for Solving Linear Equation

Variable b cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 8bx^{2}, the least common multiple of 8,bx^{2}.

$$bx^{2}\sqrt{2}=8\sqrt{x-a}$$

Reorder the terms.

$$\sqrt{2}bx^{2}=8\sqrt{x-a}$$

The equation is in standard form.

$$\sqrt{2}x^{2}b=8\sqrt{x-a}$$

Divide both sides by \sqrt{2}x^{2}.

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

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

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

Divide 8\sqrt{x-a} by \sqrt{2}x^{2}.

$$b=\frac{4\sqrt{2x-2a}}{x^{2}}$$

Variable b cannot be equal to 0.

$$b=\frac{4\sqrt{2x-2a}}{x^{2}}\text{, }b\neq 0$$

Solve for a

$$a=-\frac{b^{2}x^{4}}{32}+x$$

$$x\neq 0\text{ and }b>0$$

Steps for Solving Radical Equations

Multiply both sides of the equation by 8bx^{2}, the least common multiple of 8,bx^{2}.

$$bx^{2}\sqrt{2}=8\sqrt{x-a}$$

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

$$8\sqrt{x-a}=bx^{2}\sqrt{2}$$

Divide both sides by 8.

$$\frac{8\sqrt{-a+x}}{8}=\frac{\sqrt{2}bx^{2}}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$\sqrt{-a+x}=\frac{\sqrt{2}bx^{2}}{8}$$

Square both sides of the equation.

$$-a+x=\frac{b^{2}x^{4}}{32}$$

Subtract x from both sides of the equation.

$$-a+x-x=\frac{b^{2}x^{4}}{32}-x$$

Subtracting x from itself leaves 0.

$$-a=\frac{b^{2}x^{4}}{32}-x$$

Divide both sides by -1.

$$\frac{-a}{-1}=\frac{\frac{b^{2}x^{4}}{32}-x}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$a=\frac{\frac{b^{2}x^{4}}{32}-x}{-1}$$

Divide -x+\frac{b^{2}x^{4}}{32} by -1.

$$a=-\frac{b^{2}x^{4}}{32}+x$$

Solve for b

$$b=\frac{4\sqrt{2\left(x-a\right)}}{x^{2}}$$

$$x>a\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Variable b cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 8bx^{2}, the least common multiple of 8,bx^{2}.

$$bx^{2}\sqrt{2}=8\sqrt{x-a}$$

Reorder the terms.

$$\sqrt{2}bx^{2}=8\sqrt{x-a}$$

The equation is in standard form.

$$\sqrt{2}x^{2}b=8\sqrt{x-a}$$

Divide both sides by \sqrt{2}x^{2}.

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

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

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

Divide 8\sqrt{x-a} by \sqrt{2}x^{2}.

$$b=\frac{4\sqrt{2x-2a}}{x^{2}}$$

Variable b cannot be equal to 0.

$$b=\frac{4\sqrt{2x-2a}}{x^{2}}\text{, }b\neq 0$$

Solve for a (complex solution)

$$a=-\frac{b^{2}x^{4}}{32}+x$$

$$arg(bx^{2})<\pi \text{ and }b\neq 0\text{ and }x\neq 0$$