256
$\begingroup$

In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use it. I have tried to figure it out by proving these two equations are equal, but I can't.

Why can I use $x = \dfrac{-b\pm \sqrt{b^{2} - 4 ac}}{2a}$ to solve all quadratic equations?

  • 13
    Look at [Wikipedia](http://en.wikipedia.org/wiki/Quadratic_equation#Derivations_of_the_quadratic_formula).2011-07-03
  • 95
    This is false. You can't use the quadratic formula to solve quadratic equations in fields of characteristic 2. But maybe you don't know what that means yet.2011-07-03
  • 28
    @KCd: Not to mention the existence of a square root.2011-07-03
  • 18
    KCd and thei mention some important conditions which become crucial in general theory. The formula requires you to be able to divide by 2 and by $a$ and to take a square root. So $b^2-4ac$, also called the Discriminant, has to be a square. "characteristic 2" is one of the settings in which you can't divide by 2. If you can't take the square root of the Discriminant there are algebraic ways of adding the square root so you can solve the equation. Likewise, if you can't divide by $a$ there are sometimes ways of extending to a situation where you can. That is where more advanced algebra begins.2011-07-03
  • 3
    ... Analysing when these things are possible leads to important insights of great mathematical significance.2011-07-03
  • 3
    @KCd: He might not know enough about math to understand complex numbers, in which case he *can't* always use the quadratic formula.2011-07-03
  • 3
    @Mark Bennet. Why can't I divide by 2?2011-07-04
  • 5
    @Lous Rhys. GF(2) is a field containing only the elements 0 and 1 with the usual addition and multiplication. You can verify yourself that it fulfills all the axioms of a field. Now, try to divide 1 by 2.2011-07-04
  • 27
    @idonno I admire your curiosity! If only precalc students were more inquisitive!2012-02-23
  • 2
    @DougSpoonwood Then it isn't a quadratic equation.2013-11-07
  • 8
    Why can we use -b/a to solve ALL linear equations of the form ax+b = 0 ?2013-12-19
  • 1
    Because it is derived from the general form $ax^2 + bx + c =0$. Maybe try substituting this into the equation?2016-12-29

21 Answers 21