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anonymous
 5 years ago
Please give an easy and fast way to solve this:
anonymous
 5 years ago
Please give an easy and fast way to solve this:

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x ^{2}xc2c ^{2}=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use the quadratic formula.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(a+b^{2}=a ^{2} + 2ab + b ^{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but isn't' a quadratic equation of the type: \[a ^{2}+2ab+b ^{2}=0\] OR \[a ^{2}2ab+b ^{2}=0\] My equation has two ve signs.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to get x = 2c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Please show the steps.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^2−xc−2*c^2=0 (x+c)(x2c)=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x = {c \pm \sqrt{c^2  4*z*(2c^2)} \over 2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^2x c2c^2=0 x=c+sqrt(c^24*1*2c^2)/2*1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0c can be treated as a constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x^{2}xcc ^{2}=x^{2}2xc+xcc ^{2}=(x2c)(x+c)=0 so we have x=2c or x=c\]
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