**Problem**: (2006-2007 Warm-Up 3 - #13) What is the area of the shape formed by connecting the solutions to $ x^3-5x+12x-7 = 0 $ in the complex plane?

**Solution**: Well, let's start off by finding those solutions. It's not hard to guess that $ x= 1 $ is a solution, so we have

$ (x-1)(x^2-4x+7) = 0 $.

Using the quadratic formula, we get the other two solutions to be

$ x = 2 \pm i \sqrt{3} $.

So our points are $ 1, 2+i\sqrt{3}, 2-i\sqrt{3} $. This makes an isosceles triangle with base $ 2 \sqrt{3} $ and height $ 1 $ in the complex plane, which has area $ \sqrt{3} $. QED.

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Comment: Pretty easy after we guessed that $ x = 1 $ was a solution; a general formula for the area probably gets ugly when the polynomial is irreducible. But in this case, we can draw the triangle and figure out the area without too many tricks.

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Practice Problem: What is the area of the shape formed by connecting the solutions to $ x^3 = 1 $ in the complex plane?

lol dat iz an equilaterial triangul

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