## Saturday, February 18, 2006

### Deriving the Quadratic Formula. Topic: Algebra. Level: AMC.

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 $x,y$.

#### 1 comment:

1. #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.