AravindG
  • AravindG
what us completing the square method..i knew it but frgt can anyone help me ?
Mathematics
jamiebookeater
  • jamiebookeater
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lgbasallote
  • lgbasallote
do you want a demonstration?
AravindG
  • AravindG
general involving a ,b c and if possible a demonstration also
AravindG
  • AravindG
it seems this method is pretty useful for solving quadratics thats why i am revising it

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lgbasallote
  • lgbasallote
firstly.... you agree that \[ax^2 + bx + c = 0\] that's the quadratic equation, yes?
AravindG
  • AravindG
yep
lgbasallote
  • lgbasallote
now..divide ALL terms by a...what do you get?
lgbasallote
  • lgbasallote
where's the x?
AravindG
  • AravindG
x^2+(b/a)x+(c/a)=0
lgbasallote
  • lgbasallote
right
lgbasallote
  • lgbasallote
now subtract c/a from both sides
AravindG
  • AravindG
\[x^2+(b/a)x=-c/a\]
lgbasallote
  • lgbasallote
good. now divide b/a by 2..what do you get?
AravindG
  • AravindG
only b/a?
lgbasallote
  • lgbasallote
yes... but don't touch the equation yet
lgbasallote
  • lgbasallote
just divide b/a by 2
AravindG
  • AravindG
\[(x+b/2a)^2=(4c-b^2)/4a\]
AravindG
  • AravindG
4c+b^2 srry
lgbasallote
  • lgbasallote
why is it - b^2
AravindG
  • AravindG
4c+b^2 srry
lgbasallote
  • lgbasallote
\[x^2 + (\frac ba)x + \frac{b^2}{4a^2} = -\frac ca + \frac{b^2}{4a^2}\] \[\implies (x + \frac ba)^2 = \frac{b^2 - 4ac}{4a^2}\] you rushed too fast
AravindG
  • AravindG
i see
AravindG
  • AravindG
thx
lgbasallote
  • lgbasallote
now take the square root of both sides
lgbasallote
  • lgbasallote
do you need further help?
AravindG
  • AravindG
nop gt it
lgbasallote
  • lgbasallote
okay then
AravindG
  • AravindG
@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!"
lgbasallote
  • lgbasallote
you can use completing the square method anywhere

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