Solve frac{sqrt{x^2+4x+12}}{sqrt{x^2+4x+8}}=sqrt{frac{x+4/2x+8}}{y}

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

Solve for y

$$y=\frac{\sqrt{\frac{2\left(x^{2}+4x+8\right)\left(x^{2}+4x+16\right)}{x^{2}+4x+12}}}{2}$$

Steps for Solving Linear Equation

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.

$$y\left(x^{2}+4x+8\right)^{-\frac{1}{2}}\sqrt{x^{2}+4x+12}=\sqrt{\frac{x+4}{2}x+8}$$

Express \frac{x+4}{2}x as a single fraction.

$$y\left(x^{2}+4x+8\right)^{-\frac{1}{2}}\sqrt{x^{2}+4x+12}=\sqrt{\frac{\left(x+4\right)x}{2}+8}$$

Use the distributive property to multiply x+4 by x.

$$y\left(x^{2}+4x+8\right)^{-\frac{1}{2}}\sqrt{x^{2}+4x+12}=\sqrt{\frac{x^{2}+4x}{2}+8}$$

Divide each term of x^{2}+4x by 2 to get \frac{1}{2}x^{2}+2x.

$$y\left(x^{2}+4x+8\right)^{-\frac{1}{2}}\sqrt{x^{2}+4x+12}=\sqrt{\frac{1}{2}x^{2}+2x+8}$$

The equation is in standard form.

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

Divide both sides by \left(x^{2}+4x+8\right)^{-\frac{1}{2}}\sqrt{x^{2}+4x+12}.

$$\frac{\frac{\sqrt{x^{2}+4x+12}}{\sqrt{x^{2}+4x+8}}y\sqrt{x^{2}+4x+8}}{\sqrt{x^{2}+4x+12}}=\frac{\sqrt{2x^{2}+8x+32}}{2\times \frac{\sqrt{x^{2}+4x+12}}{\sqrt{x^{2}+4x+8}}}$$

Dividing by \left(x^{2}+4x+8\right)^{-\frac{1}{2}}\sqrt{x^{2}+4x+12} undoes the multiplication by \left(x^{2}+4x+8\right)^{-\frac{1}{2}}\sqrt{x^{2}+4x+12}.

$$y=\frac{\sqrt{2x^{2}+8x+32}}{2\times \frac{\sqrt{x^{2}+4x+12}}{\sqrt{x^{2}+4x+8}}}$$

Divide \frac{\sqrt{2x^{2}+8x+32}}{2} by \left(x^{2}+4x+8\right)^{-\frac{1}{2}}\sqrt{x^{2}+4x+12}.

$$y=\frac{\sqrt{\frac{2\left(x^{2}+4x+8\right)\left(x^{2}+4x+16\right)}{x^{2}+4x+12}}}{2}$$

Variable y cannot be equal to 0.

$$y=\frac{\sqrt{\frac{2\left(x^{2}+4x+8\right)\left(x^{2}+4x+16\right)}{x^{2}+4x+12}}}{2}\text{, }y\neq 0$$

Solve for x

$$x=-\sqrt{\sqrt{y^{4}+16}+y^{2}-8}-2$$

$$x=\sqrt{\sqrt{y^{4}+16}+y^{2}-8}-2\text{, }y\geq \sqrt{3}$$