mathslover
  • mathslover
Hi friends this is a small tutorial regarding the proof of quadratic equation ... this is made by me
Mathematics
katieb
  • katieb
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mathslover
  • mathslover
terenzreignz
  • terenzreignz
Quadratic Formula*?
mathslover
  • mathslover
yes @terenzreignz \[\Huge{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}\]

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mathslover
  • mathslover
\[{\textbf{the general form of a quadratic equation is ax^2+bx+c=0 . .}}\] \[\textbf{it is known as quadratic equation since it's degree is 2 }\] HOW TO GET THE ROOT OF A QUADRATIC EQUATION : \[\huge{ax^2+bx+c=0}\] \[\huge{ax^2+bx=-c}\] \[\huge{\frac{(ax^2+bx)}{a}=\frac{-c}{a}}\] \[\huge{x^2+\frac{bx}{a}=\frac{-c}{a}}\] \[\huge{(x)^2+2(\frac{b}{2a})(x)=\frac{-c}{a}}\] \[\huge{(x)^2+2(\frac{b}{2a})(x)+\frac{b^2}{4a^2}=\frac{-c}{a}+\frac{b^2}{4a^2}}\] \[\huge{(x+\frac{b}{2a})^2 = \frac{-c}{a}+\frac{b^2}{4a^2}}\] \[\huge{x+\frac{b}{2a}=\pm\sqrt{\frac{-c}{a}+\frac{b^2}{4a^2}}}\] \[\huge{x=\pm\sqrt{\frac{-c}{a}+\frac{b^2}{4a^2}}-\frac{b}{2a}}\] \[\huge{x=\pm\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}}\] \[\huge{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}\]
ParthKohli
  • ParthKohli
Good work, mathslover, I appreciate your tutorial :)
mathslover
  • mathslover
thanks a lot @ParthKohli
mathslover
  • mathslover
Please prefer the LaTeX for better proof and for no mistakes
anonymous
  • anonymous
Great job @mathslover!
mathslover
  • mathslover
mathslover
  • mathslover
anonymous
  • anonymous
You are very welcome
anonymous
  • anonymous
@mathslover wat is ur intention by posting this!
mathslover
  • mathslover
to just make you all aware with quadratic equation
anonymous
  • anonymous
ok
mathslover
  • mathslover
I am going to post some more interesting things about this
anonymous
  • anonymous
plz go ahead
mathslover
  • mathslover
i will but after some time. .. doing some research work :)
anonymous
  • anonymous
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mathslover
  • mathslover
@wdkekefkjrgjr dont post such comments
mathslover
  • mathslover
i m posting more interesting things about quadratic equation as a new question
jiteshmeghwal9
  • jiteshmeghwal9
k! thanx a lot @mathslover :)
mathslover
  • mathslover
Ur welcome @jiteshmeghwal9 Best of Luck

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