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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alternative, formula, quadratic 
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