Quadratic formula: Difference between revisions

Latest revision as of 23:41, 1 July 2012

Statement

Consider a quadratic equation of the form:

$ax^{2}+bx+c=0,a,b,c\in \mathbb {R} ,a\neq 0$

This is treated as an equation in $x$ . The quadratic formula is a formula to determine the solutions of this equation. The short version of the formula is that the roots are:

${\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$

if the expression makes sens; otherwise there are no roots.

To understand the formula, first define the discriminant of the quadratic function $ax^{2}+bx+c$ as the value $b^{2}-4ac$ . Now, we make three cases:

Case for discriminant $b^{2}-4ac$	Conclusion for roots	Conclusion for factorization of polynomial $ax^{2}+bx+c$
positive, i.e., $b^{2}-4ac>0$	there are two real roots, given as ${\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}$ and ${\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}$	If we denote the roots by $\alpha ,\beta$ , then $ax^{2}+bx+c=a(x-\alpha )(x-\beta )$
zero, i.e., $b^{2}-4ac=0$	there is a single real root with multiplicity two, and that root is $-b/2a$	$ax^{2}+bx+c=a(x+(b/2a))^{2}$
negative, i.e., $b^{2}-4ac<0$	there are no real roots	the polynomial does not factor, i.e., it is irreducible

@@ Line 16: / Line 16: @@
 ! Case for discriminant <math>b^2 - 4ac</math> !! Conclusion for roots !! Conclusion for factorization of polynomial <math>ax^2 + bx + c</math>
 |-
-| positive, i.e., <math>b^2 - 4ac > 0</math> || there are two real roots, given as <math>\frac{-b + \sqrt{b^2 - 4ac}}{2a}</math> and <math><math>\frac{-b - \sqrt{b^2 - 4ac}}{2a}</math> || If we denote the roots by <math>\alpha, \beta</math>, then <math>ax^2 + bx + c = a(x- \alpha)(x - \beta)</math>
+| positive, i.e., <math>b^2 - 4ac > 0</math> || there are two real roots, given as <math>\frac{-b + \sqrt{b^2 - 4ac}}{2a}</math> and <math>\frac{-b - \sqrt{b^2 - 4ac}}{2a}</math> || If we denote the roots by <math>\alpha, \beta</math>, then <math>ax^2 + bx + c = a(x- \alpha)(x - \beta)</math>
 |-
-| zero, i.e., <math>b^2 - 4ac = 0</math> || there is a single real root with multiplicity two, and that root is <math>-b/2a</mah> || <math>ax^2 + bx + c = a(x + (b/2a))^2</math>
+| zero, i.e., <math>b^2 - 4ac = 0</math> || there is a single real root with multiplicity two, and that root is <math>-b/2a</math> || <math>ax^2 + bx + c = a(x + (b/2a))^2</math>
 |-
 | negative, i.e., <math>b^2 - 4ac < 0</math> || there are no real roots || the polynomial does not factor, i.e., it is irreducible
 |}