# An underdetermined linear system for GPS

I'm reading the paper "An Underdetermined Linear System for GPS" by Dan Kalman, and solving the equations. It's all going fine until near the bottom of page 388, where the equation $$0.02t^2-1.88t+43.56=0$$ is stated and subsequently solved:

[...] leading to two solutions, 43.1 and 50.0. If we select the first solution, then $(x, y, z) = (1.317, 1.317, 0.790)$, which has a length of about $2$. We are using units of earth radii, so this point is around 4000 miles above the surface of the earth. The second value of $t$ leads to $(x, y, z) = (.667, .667, .332)$, with length $0.9997$. That places the point on the surface of the earth (to four decimal places) and gives us the location of the ship.

My question is how he got the values 43.1 and 50.0? Every time I solve it, using the quadratic formula, I get 41.4 and 52.5, which is different.

Here's the link for the paper (it's only 7 pages long): http://mathdl.maa.org/images/upload_library/22/Polya/Kalman.pdf

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## 1 Answer

Above your given equation there is another one, not simplified $$(5.41 − 0.095t − 1)^2 + (5.41 − 0.095t − 2)^2 + (3.67 − 0.067t)^2 = 0.047^2(t − 19.9)^2$$ which gives more accurate form $$0.02033^2-1.8896618t+43.67031391=0$$ and it has roots $49.913...$, $43.035...$. So you shouldn't worry, all these errors are because of rounding

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thanks very much – HATEM EL-AZAB Apr 15 '12 at 15:20