Proof of the Quadratic formula

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The Quadratic Formula used to solve equations of the form ax^2+bx+c=0 \,\!. It is:

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


Dividing our quadratic equation by a\,\! (which is allowed because a\,\! is non-zero), we have

x^2 + \frac{b}{a}  x + \frac{c}{a}=0

which is equivalent to

x^2 + \frac{b}{a} x= -\frac{c}{a}.

The equation is now in a form in which we can conveniently complete the square. To "complete the square" is to add a constant (i.e., in this case, a quantity that does not depend on x\,\!) to the expression to the left of "=\,\!", that will make it a perfect square trinomial of the form x^2+2xy+y^2\,\!. Since 2xy\,\! in this case is \frac{b}{a}  x , we must have y = \frac{b}{2a}, so we add the square of \frac{b}{2a} to both sides, getting


The left side is now a perfect square; it is the square of \left(x + \frac{b}{2a}\right). The right side can be written as a single fraction; the common denominator is 4a^2\,\!. We get


Taking square roots of both sides yields

\left|x+\frac{b}{2a}\right| = \frac{\sqrt{b^2-4ac\  }}{|2a|}\Leftrightarrowx+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac\  }}{2a}

Subtracting \frac{b}{2a} from both sides, we get

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

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