Need help with algebra variation word problem I've been trying to re-learn algebra by using the book called Beginning and Intermediate Algebra by Tyler Wallace and I have been stuck on variation word problems. Most of them I'm able kinda solve after breaking it down but there's this one which I been stuck on.
The weight of an object varies inversely as the square of the distance from the
center of the earth. At sea level (6400 km from the center of the earth), an
astronaut weighs 100 lb. How far above the earth must the astronaut be in
order to weigh 64 lb?

The formulas I have tried are 6400^2(100)/64, 100^2(6400)/64 and 6400(100)/64 and also tried to square root the answers to see if it matches the answers given in the book which is 1600km. I appreciate any hints or explanation, thanks in advance.
 A: The weight varies inversely with some quantity means that the weight is proportional to the inverse of that quantity. Since the weight varies inversely with the square of the distance we choose a variable for the distance, say $r$. Then the weight $w$ is inversely proportional to $\frac{1}{r^2}$. When a variable is proportional to another, it means that it is equal to that variable times a constant, call it $k$. Then we get that $$w=\frac{k}{r^2}$$
Now use the given information to solve for $k$. You will then have a complete formula that you can use to solve the second part of the problem.
A: We isolate mass, equate that mass, and solve the quadratic for a positive solution
\begin{align*}
\dfrac{m}{6400^2}=100    &\implies m=100(6400)^2\\
\dfrac{m}{(6400+x)^2}=64 &\implies m=64(6400+x)^2\\ \\
100(6400)^2&=64(6400+x)^2\\
0&=64(6400+x)^2-100(6400)^2\\
64 x^2 + 819200 x - 1474560000 &= 0\\
64 (x - 1600) (x + 14400) &= 0\\
x=1600
\end{align*}
Wolfram Alpha handles much of the tedium of expanding, simplifying and factoring as shown here.
