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 August 24th, 2017, 12:25 AM #41 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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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