[tex]\text{Solve for x:}\\\\\frac{3}{x}=\frac{x}{12}\\\\\text{Multiply both sides by x}\\\\3=\frac{x}{12}x\\\\\text{Multiply both sides by 12}\\\\36=x\cdot x\\\\36=x^2\\\\\text{To format it, subtract 36 and}\,\,x^2\text{ on both sides}\\\\-x^2=-36\\\\\text{Divide both sides by -1}\\\\x^2=36\\\\\text{Square root:}\\\\\sqrt x^2=\pm\sqrt36\\\\\boxed{x=6\,\,or\,\,x=-6}[/tex]
Answer:
x = ± 6
Step-by-step explanation:
Given
[tex]\frac{3}{x}[/tex] = [tex]\frac{x}{12}[/tex] ( cross- multiply )
x² = 36 ( take the square root of both sides )
[tex]\sqrt{x^2}[/tex] = ± [tex]\sqrt{36}[/tex], that is
x = ± 6
