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as i'm reading a paper An Underdetermined Linear System for GPS By Dan Kalman

and solving the equations it's all going fine until in page 388,down the page ,it's wrote according to this equation $$0.02t2-1.88t+43.56=0$$ that this quote,

"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?that every time i solve it using quadratic formula i got 41.4 and 52.5 which is different

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Right, this is high school algebra. –  duffymo Apr 15 '12 at 16:12

1 Answer 1

up vote 1 down vote accepted

The answers 43.1 and 50.0 are correct answers to the previous equation:

(5.41 − .095t − 1)^2 + (5.41 − .095t − 2)^2 + (3.67 − .067t)^2
= .047^2(t − 19.9)^2

The quadratic equation is meant to be an expansion of this equation.

The correct expansion is given by:

43.67031391 - 1.8896618*x + 0.02033*x**2

which still has the same answers.

However, the text contains a slightly less accurate expansion (which is actually given if you replace 0.047 with 0.05 before you expand the original equation) which is why it has different solutions.

My guess is that the author solved it at high precision, and then when writing the paper felt that an intermediate step was justified but did not use the same degree of precision when computing the simplifed quadratic equation.

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yah you're 100% correct...thanks very much –  HATEM EL-AZAB Apr 16 '12 at 5:32

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