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mathslover

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

  • one year ago
  • one year ago

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  1. mathslover
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    • one year ago
  2. terenzreignz
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    Quadratic Formula*?

    • one year ago
  3. mathslover
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    yes @terenzreignz \[\Huge{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}\]

    • one year ago
  4. mathslover
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    \[{\textbf{the general form of a quadratic equation is ax^2+bx+c=0 . .}}\] \[\textbf{it is known as quadratic equation since it's degree is 2 }\] HOW TO GET THE ROOT OF A QUADRATIC EQUATION : \[\huge{ax^2+bx+c=0}\] \[\huge{ax^2+bx=-c}\] \[\huge{\frac{(ax^2+bx)}{a}=\frac{-c}{a}}\] \[\huge{x^2+\frac{bx}{a}=\frac{-c}{a}}\] \[\huge{(x)^2+2(\frac{b}{2a})(x)=\frac{-c}{a}}\] \[\huge{(x)^2+2(\frac{b}{2a})(x)+\frac{b^2}{4a^2}=\frac{-c}{a}+\frac{b^2}{4a^2}}\] \[\huge{(x+\frac{b}{2a})^2 = \frac{-c}{a}+\frac{b^2}{4a^2}}\] \[\huge{x+\frac{b}{2a}=\pm\sqrt{\frac{-c}{a}+\frac{b^2}{4a^2}}}\] \[\huge{x=\pm\sqrt{\frac{-c}{a}+\frac{b^2}{4a^2}}-\frac{b}{2a}}\] \[\huge{x=\pm\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}}\] \[\huge{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}\]

    • one year ago
  5. ParthKohli
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    Good work, mathslover, I appreciate your tutorial :)

    • one year ago
  6. mathslover
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    thanks a lot @ParthKohli

    • one year ago
  7. mathslover
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    Please prefer the LaTeX for better proof and for no mistakes

    • one year ago
  8. rebeccaskell94
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    Great job @mathslover!

    • one year ago
  9. mathslover
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    hanks @rebeccaskell94

    • one year ago
  10. mathslover
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    *thanks @rebeccaskell94

    • one year ago
  11. rebeccaskell94
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    You are very welcome

    • one year ago
  12. Yahoo!
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    @mathslover wat is ur intention by posting this!

    • one year ago
  13. mathslover
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    to just make you all aware with quadratic equation

    • one year ago
  14. Yahoo!
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    ok

    • one year ago
  15. mathslover
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    I am going to post some more interesting things about this

    • one year ago
  16. Yahoo!
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    plz go ahead

    • one year ago
  17. mathslover
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    i will but after some time. .. doing some research work :)

    • one year ago
  18. wdkekefkjrgjr
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    fhjjfjfjjfjgkhkjkjhk h kh khkghkghkhhkgffgfg

    • one year ago
  19. mathslover
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    @wdkekefkjrgjr dont post such comments

    • one year ago
  20. mathslover
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    i m posting more interesting things about quadratic equation as a new question

    • one year ago
  21. mathslover
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    http://openstudy.com/study#/updates/4fe09fa0e4b06e92b86fbb47

    • one year ago
  22. jiteshmeghwal9
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    k! thanx a lot @mathslover :)

    • one year ago
  23. mathslover
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    Ur welcome @jiteshmeghwal9 Best of Luck

    • one year ago
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