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Mathematics 8 Online
OpenStudy (anonymous):

help please x^2+2x+1-Q^2=

OpenStudy (alfie):

If Q is a numeric constant, you can solve the equation with the quadratic formula but setting a = 1 b = 2 c = +1-q^2 Of course the delta won't give you a perfect root, and you'll have some solutions where this Q appears everywhere.

OpenStudy (anonymous):

thnks :O)

OpenStudy (alfie):

Or, another way you could do it, is applying the following: (a+b)^2 = a^2+2ab+b^2 So... (x+1)^2 - Q^2 Now. \[a^2 - b^2 = (a+b)(a-b)\] So \[(x+1-Q)(x+1+Q)\] = 0 Zero product property and... x = Q-1 x = -1-Q

OpenStudy (anonymous):

that makes it much easier thank you!

OpenStudy (alfie):

You're welcome!

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