Proving/Deriving the Quadratic EquationDate: 4/1/96 at 12:27:1 From: Anonymous Subject: Quadratic equation solution How can one derive the quadratic equation, i.e. prove it? I memorized it in middle school along with every other seventh grader, but I don't remember ever seeing any sort of analysis or proof. How did someone come up with it? Date: 4/4/96 at 2:58:28 From: Doctor Aaron Subject: Re: Quadratic equation solution Hi, Good question, I love to answer this one. Do you remember how you solved quadratics before you memorized the formula? Its a technique called completing the square. That is (x + E)^2 = x^2 +2Ex +E^2. The quadratic formula is just getting an arbitrary quadratic, Ax^2 + Bx + C = 0 into this format by adding and subtracting some stuff, then we get a big mess on the other side, but we have a perfect square so we can take its square root. Lets do it. Let A,B,C be real numbers and let Ax^2 + Bx + C = 0 First we get rid of the A by dividing through x^2 + (B/A)x + (C/A) = 0 Now (B/A) must equal 2*E if we are going to get a perfect square so we want x^2 + (B/A)x + (B/(2A))^2 = something but something is just what we get when we add (B/(2A))^2 and subtract (C/A) from both sides. then x^2 + (B/A)x + (B/(2A))^2 = (B/(2A))^2 - (C/A) we can multiply the right side by 4A/4A to get B^2 4AC B^2 - 4AC --- - --- = ----------- on the right. 4A^2 4A^2 4A^2 We have a perfect square on the left, so taking the square root of both sides, we get: +- sqrt (B^2 - 4AC) +- sqrt (B^2 - 4AC) x + B/(2A) = --------------------- = ---------------------. sqrt (4A^2) 2A Subtracting B/(2A) from both sides we get the familiar x = -B (+ or -) sqrt (B^2 - 4AC) ------------------------------ 2A. -Doctor Aaron, The Math Forum |
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