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what us completing the square method..i knew it but frgt can anyone help me ?
 one year ago
 one year ago
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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lgbasalloteBest ResponseYou've already chosen the best response.2
do you want a demonstration?
 one year ago

AravindGBest ResponseYou've already chosen the best response.1
general involving a ,b c and if possible a demonstration also
 one year ago

AravindGBest ResponseYou've already chosen the best response.1
it seems this method is pretty useful for solving quadratics thats why i am revising it
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
firstly.... you agree that \[ax^2 + bx + c = 0\] that's the quadratic equation, yes?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
now..divide ALL terms by a...what do you get?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
now subtract c/a from both sides
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
good. now divide b/a by 2..what do you get?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
yes... but don't touch the equation yet
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
just divide b/a by 2
 one year ago

AravindGBest ResponseYou've already chosen the best response.1
\[(x+b/2a)^2=(4cb^2)/4a\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
\[x^2 + (\frac ba)x + \frac{b^2}{4a^2} = \frac ca + \frac{b^2}{4a^2}\] \[\implies (x + \frac ba)^2 = \frac{b^2  4ac}{4a^2}\] you rushed too fast
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
now take the square root of both sides
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
do you need further help?
 one year ago

AravindGBest ResponseYou've already chosen the best response.1
@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

lgbasalloteBest ResponseYou've already chosen the best response.2
you can use completing the square method anywhere
 one year ago
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