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AravindG

  • 2 years ago

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

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  1. lgbasallote
    • 2 years ago
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    do you want a demonstration?

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

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

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

  5. AravindG
    • 2 years ago
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    yep

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

  7. lgbasallote
    • 2 years ago
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    where's the x?

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

  9. lgbasallote
    • 2 years ago
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    right

  10. lgbasallote
    • 2 years ago
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    now subtract c/a from both sides

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

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

  13. AravindG
    • 2 years ago
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    only b/a?

  14. lgbasallote
    • 2 years ago
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    yes... but don't touch the equation yet

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

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

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

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

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

  20. lgbasallote
    • 2 years ago
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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

  21. AravindG
    • 2 years ago
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    i see

  22. AravindG
    • 2 years ago
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    thx

  23. lgbasallote
    • 2 years ago
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    now take the square root of both sides

  24. lgbasallote
    • 2 years ago
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    do you need further help?

  25. AravindG
    • 2 years ago
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    nop gt it

  26. lgbasallote
    • 2 years ago
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    okay then

  27. AravindG
    • 2 years ago
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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!"

  28. lgbasallote
    • 2 years ago
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    you can use completing the square method anywhere

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