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August 24th, 2017, 01:25 AM   #41
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Well suppose

$x^2 = 10^{1000}x + 1$

We can do

$ x = \frac{(10^{1000}) \pm \sqrt{(10^{1000})^2 + 4(1)(1)}}{2(1)} $

$ x = \frac{(10^{1000}) \pm \sqrt{10^{2000} + 4}}{2} $


The solution we get by adding the radical is very close to $ \ \ x = 10^{1000} \ \ $

Obviously $ \ \ x = 0 \ \ $ is not a solution , if we want accuracy it is going to be difficult to calculate this radical.

How will putting the radical in the denominator make our work any easier?
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