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## Homework Statement

[tex]\left(\frac{\dot{x}}{a}\right)^2 = K\left[b\frac{a}{x} + c\left(\frac{a}{x}\right)^2 + (1-b-c)\right][/tex]

Show that, for [itex]b<1[/itex], there is a value of [itex]c[/itex] that makes the right hand side a perfect square of a function of x.

**2. The attempt at a solution**

I guess that a perfect square is a square of a function that can be written as

[tex](x \pm d)^2 = x^2 \pm 2xd + d^2[/tex]

but I have not been successful in this problem.