Derivation of Quadratic FormulaDate: 6/12/96 at 20:46:1 From: Milton F. Lopez Subject: Derivation of quadratic formula Confession: my K-12 days are long gone ... Request: I would like to derive the solution to ax(2) + bx + c = 0. I did this years ago in a college level analysis class, but I've lost a lot of neurons since then. Can you help? Most places I've looked simply state the soluion without proof. Thanks in advance. Date: 6/13/96 at 20:54:32 From: Doctor Ceeks Subject: Re: Derivation of quadratic formula ax^2+bx+c = a(x+b/(2a))^2-b^2/(4a)+c = 0, so that (x+b/(2a))^2=b^2/(4a^2)-c/a. Now take square roots and subtract b/(2a) from both sides. -Doctor Ceeks, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 6/13/96 at 20:52:30 From: Doctor Luis Subject: Re: Derivation of quadratic formula In order to find a solution to the general quadratic equation ax^2+bx+c=0 one needs to "complete the square" on the terms ax^2+bx. By doing this you are transforming the general quadratic into the form (x+m)^2-n=0, which we know has the solution x= -m +/- sqrt(n). ("+/-" plus or minus, and sqrt() is the square root function) Okay? Claim: The solutions to the general quadratic ax^2+bx+c=0 are (-b+sqrt(b^2-4ac))/2a and (-b-sqrt(b^2-4ac))/2a Proof: Consider the quadratic equation: ax^2+bx+c=0 (General form) Now divide by a (a/a)x^2+(b/a)x+(c/a)=0/a x^2+(b/a)x+(c/a)=0 Add and subtract (b/2a)^2 (a form of 0) x^2+(b/a)x+(b/2a)^2-(b/2a)^2+(c/a)=0 Now, examine very closely the parenthesized expression. It looks like the perfect square of a binomial (and it is) [x^2+(b/a)x+(b/2a)^2] - (b/2a)^2+(c/a)=0 [x^2+2(b/2a)x+(b/2a)^2] - (b/2a)^2+(c/a)=0 If you let m = b/2a, then [x^2+2mx+m^2] - (b/2a)^2+(c/a)=0 [x+m]^2 - (b/2a)^2+(c/a)=0 Okay. So far we have: (x+m)^2 + (c/a)-(b/2a)^2 = 0 (x+m)^2 + (c/a)-(b^2/4a^2) = 0 (x+m)^2 + (4ac-b^2)/(4a^2) = 0 (x+m)^2 = (b^2-4ac)/(4a^2) (x+m) = (+/-)sqrt(b^2-4ac)/sqrt(4a^2) (x+m) = (+/-)sqrt(b^2-4ac)/2a x = -m +/- sqrt(b^2-4ac)/2a x = -b/2a +/- sqrt(b^2-4ac)/2a x = (-b +/- sqrt(b^2-4ac))/2a Quod erat demonstrandum [Latin for "which was to be proven"] Right now the proof might look complicated, but if you write it on paper it looks clearer than with ASCII. -Doctor Luis, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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