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AravindG

what us completing the square method..i knew it but frgt can anyone help me ?

  • one year ago
  • one year ago

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  1. lgbasallote
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    do you want a demonstration?

    • one year ago
  2. AravindG
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    general involving a ,b c and if possible a demonstration also

    • one year ago
  3. AravindG
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    it seems this method is pretty useful for solving quadratics thats why i am revising it

    • one year ago
  4. lgbasallote
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    firstly.... you agree that \[ax^2 + bx + c = 0\] that's the quadratic equation, yes?

    • one year ago
  5. AravindG
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    yep

    • one year ago
  6. lgbasallote
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    now..divide ALL terms by a...what do you get?

    • one year ago
  7. lgbasallote
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    where's the x?

    • one year ago
  8. AravindG
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    x^2+(b/a)x+(c/a)=0

    • one year ago
  9. lgbasallote
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    right

    • one year ago
  10. lgbasallote
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    now subtract c/a from both sides

    • one year ago
  11. AravindG
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    \[x^2+(b/a)x=-c/a\]

    • one year ago
  12. lgbasallote
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    good. now divide b/a by 2..what do you get?

    • one year ago
  13. AravindG
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    only b/a?

    • one year ago
  14. lgbasallote
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    yes... but don't touch the equation yet

    • one year ago
  15. lgbasallote
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    just divide b/a by 2

    • one year ago
  16. AravindG
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    \[(x+b/2a)^2=(4c-b^2)/4a\]

    • one year ago
  17. AravindG
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    4c+b^2 srry

    • one year ago
  18. lgbasallote
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    why is it - b^2

    • one year ago
  19. AravindG
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    4c+b^2 srry

    • one year ago
  20. lgbasallote
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    \[x^2 + (\frac ba)x + \frac{b^2}{4a^2} = -\frac ca + \frac{b^2}{4a^2}\] \[\implies (x + \frac ba)^2 = \frac{b^2 - 4ac}{4a^2}\] you rushed too fast

    • one year ago
  21. AravindG
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    i see

    • one year ago
  22. AravindG
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    thx

    • one year ago
  23. lgbasallote
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    now take the square root of both sides

    • one year ago
  24. lgbasallote
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    do you need further help?

    • one year ago
  25. AravindG
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    nop gt it

    • one year ago
  26. lgbasallote
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    okay then

    • one year ago
  27. AravindG
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    @lgbasallote can you tell me how we can say that this method will be easier when we have an arbitrary quadratic?i mean is there any relation btw a,b,c so that i can recognise "AHA I CAN USE COMPLETING THE SQUARE HERE INSTEAD OF OTHER METHODS!"

    • one year ago
  28. lgbasallote
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    you can use completing the square method anywhere

    • one year ago
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