Expert: Scott A Wilson - 3/13/2011

Question
using the method to solve quadratic equations that came from India how would i solve this:
                            x^-2x-13=0
i understand the first few steps:
x^-2x=13
4x^-8x=52
4x^-8x+4=52+4
4x^-8x+4=56
how do i solve for x now?

Answer
If you got 4x^2 - 8x + 4 = 56, I would factor that into (2x-2)(2x-2) = 56.
It could then be said that 2x-2 = ±√56 = ±2√14, so x-1 = ±√14, , so x = 1±√14.

If you went back and multiplied (x-1+√14)(x-1-√14), you would be x² - 2x - (14-1) = x² - 2x - 13.

I have never seen the method from India, but I'm not sure why the method multiplies by 4.
It would work just as well to say that x²-2x+1 = 13+1, so (x-1)² = 14.