Question

U can use this formula to find the value of 'x' in Quadratic Equations
It's so useful

Answer 1

\[\Huge{\color{red}{\frac{-b \pm \sqrt{b^2-4ac}}{2a}}}\]

Answer 2

use this formula for equations like \[ax^2\pm bx \pm c\]

Answer 3

cause you the viper
http://www.youtube.com/watch?v=TKFH_zh4gY0

Answer 4

is there a question with this?

Answer 5

e.g, 5x^2-3x-1
here a=5, b=-3 & c=-1

Answer 6

right

Answer 7

Putting these values in formula, we will get two values.

Answer 8

for x*

Answer 9

yes - in this case you will get 2 values for x

Answer 10

first, \[x=\frac{-5+\sqrt{29}}{10}\]
second, \[x=\frac{-5-\sqrt{29}}{10}\]

Answer 11

no

its -b in the numerator - that is -(-3) = 3

Answer 12

not -5

Answer 13

ty lol

Answer 14

yes u got that I was trying if anyone could catch this before giving me a medal haha & u deserved a medal:)

Answer 15

so\[x=\frac{3+\sqrt{29}}{10}\]or\[x=\frac{3-\sqrt{29}}{10}\]

Answer 16

yea

Answer 17

haha thanx sir @satellite73 :)

Answer 18

@satellite73 sir plz plz plz change ur pic to this:-
http://upload.wikimedia.org/wikipedia/commons/8/8d/GPS_Satellite_NASA_art-iif.jpg

Answer 19

it will match ur name @satellite73

Answer 20

@TheViper, I'm glad you wanted to share your joy with the quadratic formula. :)

Answer 21

thanx @Limitless

Answer 22

the quadratic formula is very useful :)

Answer 23

yes right

Answer 24

Anyone up for the Cubic Formula (for a leading coefficient of 1)?
\[Ax^3+Bx^2+Cx+D=0\]
Let \(a=\frac{B}{A}\), \(b=\frac{C}{A}\), and \(c=\frac{D}{A}.\) Then,

\[
\begin{eqnarray*}
r_1 & = & -\frac{a}{3} + \left(\frac{-2a^3 + 9ab - 27c + \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\
& & {} + \left(\frac{-2a^3 + 9ab - 27c - \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\
r_2 & = & -\frac{a}{3} - \frac{1+i\sqrt{3}}{2} \left(\frac{-2a^3 + 9ab - 27c + \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\
& & {} + \frac{-1+i\sqrt{3}}{2} \left(\frac{-2a^3 + 9ab - 27c - \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\
r_3 & = & -\frac{a}{3} + \frac{-1+i\sqrt{3}}{2} \left(\frac{-2a^3 + 9ab - 27c + \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\
& & {} - \frac{1+i\sqrt{3}}{2} \left(\frac{-2a^3 + 9ab - 27c - \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3}.
\end{eqnarray*}
\]

Answer 25

wow

Answer 26

thanx

Answer 27

source: http://planetmath.org/?method=src&id=1407&op=getobj&from=objects

Answer 28

no limit to give answers

Answer 29

lol @Limitless =))))