Vipul: /* Statement */

2012-07-01T23:41:09Z

Statement

← Older revision		Revision as of 23:41, 1 July 2012
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	! Case for discriminant <math>b^2 - 4ac</math> !! Conclusion for roots !! Conclusion for factorization of polynomial <math>ax^2 + bx + c</math>		! Case for discriminant <math>b^2 - 4ac</math> !! Conclusion for roots !! Conclusion for factorization of polynomial <math>ax^2 + bx + c</math>
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	\| 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		\| negative, i.e., <math>b^2 - 4ac < 0</math> \|\| there are no real roots \|\| the polynomial does not factor, i.e., it is irreducible
	\|}		\|}

Vipul: Created page with "==Statement== Consider a quadratic equation of the form: $ax^2 + bx + c = 0, a,b,c \in \R, a \ne 0$ This is treated as an equation in $x$ . The '''quad..."

2012-07-01T23:22:14Z

Created page with "==Statement== Consider a quadratic equation of the form: <math>ax^2 + bx + c = 0, a,b,c \in \R, a \ne 0</math> This is treated as an equation in <math>x</math>. The '''quad..."

New page

==Statement==

Consider a quadratic equation of the form:

<math>ax^2 + bx + c = 0, a,b,c \in \R, a \ne 0</math>

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

<math>\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}</math>

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

To understand the formula, first define the [[discriminant]] of the quadratic function <math>ax^2 + bx + c</math> as the value <math>b^2 - 4ac</math>. Now, we make three cases:

{| class="sortable" border="1"
! 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>
|-
| 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>
|-
| negative, i.e., <math>b^2 - 4ac < 0</math> || there are no real roots || the polynomial does not factor, i.e., it is irreducible
|}

Quadratic formula - Revision history

Vipul: /* Statement */

Vipul: Created page with "==Statement== Consider a quadratic equation of the form: ax^2 + bx + c = 0, a,b,c \in \R, a \ne 0 This is treated as an equation in x. The '''quad..."

Vipul: Created page with "==Statement== Consider a quadratic equation of the form: $ax^2 + bx + c = 0, a,b,c \in \R, a \ne 0$ This is treated as an equation in $x$ . The '''quad..."