Question

Hi friends this is a small tutorial regarding the proof of quadratic equation ... this is made by me

Answer 1

Quadratic Formula*?

Answer 2

yes @terenzreignz
\[\Huge{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}\]

Answer 3

\[{\textbf{the general form of a quadratic equation is ax^2+bx+c=0 . .}}\]
\[\textbf{it is known as quadratic equation since it's degree is 2 }\]

HOW TO GET THE ROOT OF A QUADRATIC EQUATION :

\[\huge{ax^2+bx+c=0}\]
\[\huge{ax^2+bx=-c}\]
\[\huge{\frac{(ax^2+bx)}{a}=\frac{-c}{a}}\]
\[\huge{x^2+\frac{bx}{a}=\frac{-c}{a}}\]
\[\huge{(x)^2+2(\frac{b}{2a})(x)=\frac{-c}{a}}\]
\[\huge{(x)^2+2(\frac{b}{2a})(x)+\frac{b^2}{4a^2}=\frac{-c}{a}+\frac{b^2}{4a^2}}\]
\[\huge{(x+\frac{b}{2a})^2 = \frac{-c}{a}+\frac{b^2}{4a^2}}\]
\[\huge{x+\frac{b}{2a}=\pm\sqrt{\frac{-c}{a}+\frac{b^2}{4a^2}}}\]
\[\huge{x=\pm\sqrt{\frac{-c}{a}+\frac{b^2}{4a^2}}-\frac{b}{2a}}\]
\[\huge{x=\pm\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}}\]
\[\huge{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}\]

Answer 4

Good work, mathslover, I appreciate your tutorial :)

Answer 5

thanks a lot @ParthKohli

Answer 6

Please prefer the LaTeX for better proof and for no mistakes

Answer 7

Great job @mathslover!

Answer 8

hanks @rebeccaskell94

Answer 9

*thanks @rebeccaskell94

Answer 10

You are very welcome

Answer 11

@mathslover wat is ur intention by posting this!

Answer 12

to just make you all aware with quadratic equation

Answer 13

ok

Answer 14

I am going to post some more interesting things about this

Answer 15

plz go ahead

Answer 16

i will but after some time. .. doing some research work :)

Answer 17

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Answer 18

@wdkekefkjrgjr dont post such comments

Answer 19

i m posting more interesting things about quadratic equation as a new question

Answer 20

http://openstudy.com/study#/updates/4fe09fa0e4b06e92b86fbb47

Answer 21

k! thanx a lot @mathslover :)

Answer 22

Ur welcome @jiteshmeghwal9

Best of Luck