**Problem**: (Quadratic Formula) Given positive reals $ a,b,c $, find a formula in terms of $ a,b,c $ that gives the roots of the quadratic $ ax^2+bx+c = 0 $.

**Solution**: The formula is obviously very well-known, but the derivation is interesting and is an interesting use of a method called

*completing the square*.

From $ ax^2+bx+c = 0 $ we first divide through by $ a $ to get a monic quadratic. This makes completing the square easier. So we have $ x^2+\frac{b}{a}x+\frac{c}{a} = 0 $.

As completing the square implies, we want to make a square. It is important that both the $ x^2 $ and $ x $ terms are in the square so we have only a constant remaining. We rewrite our quadratic as

$ \left(x+\frac{b}{2a}\right)^2+\left(\frac{c}{a}-\frac{b^2}{4a^2}\right) = 0 $, or

$ \left(x+\frac{b}{2a}\right)^2 = \frac{b^2-4ac}{4a^2} $.

From here we see the seemingly random terms of the quadratic formula appear. Square rooting both sides, we find

$ x+\frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{2a} $, and

$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $

as desired. QED.

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Comment: The solution is quite simple, and the only key step is completing the square, which pretty much finished it off.

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Practice Problem #1: Graph $ 3x^2+18x+5y^2+20y = 96 $.

Practice Problem #2: Solve $ x^2+2x = 4y+2 $ for positive integers $ x,y $.

#2: Adding 1 to both sides of the equality gives us $(x+1)^2=4y+3\equiv 3 \pmod {4}$, so there are no solutions.

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