Home > CALC > Chapter 6 > Lesson 6.1.4 > Problem6-52

6-52.

There are many ways to solve this equation.

$\text{One method is to let }U=\frac{1}{x}\text{ and solve for }U\text{ first.}$

$\frac{5U^2}{1+25U^2}=\frac{U^2}{1+U^2}$

$5U^2+5U^4=U^2+25U^4$

$4U^2-20U^4=0$

$4U^2(1-5U^2)=0$