Mathematics 91 Online OpenStudy (anonymous):

Four different methods of solving a quadratic equation have been discussed in this course: factoring, the square root property, completing the square, and the quadratic formula. Explain under what circumstances each method would be preferred over any of the other methods. Give an example for each circumstance. OpenStudy (saifoo.khan):

Discussion Question? OpenStudy (anonymous):

Yes... I don't normally need help with these... this is a very tough lesson week :( OpenStudy (saifoo.khan):

Factoring should be done where the equation is Factorable. Like: x^2-x+6=0 in this equation the factors are (x+2)(x-3) OpenStudy (anonymous):

you need the formula when you can't factorize it. you should to factorize it because it's the easiest. you need to complete the square when you need to put in the general formula : \[a (x-h)^2 +k\] i have no idea what is the square root property. OpenStudy (dumbcow):

If the quadratic has no middle term, then use square root property: x^2 -8 = 0 --> x^2 = 8 --> x = +-sqrt(8) If there are factors of a*c which add up to b. Then it can be factored x^2 -4x -12 =0 --> -6*2 = 12 and -6+2 = -4 --> (x-6)(x+2)=0 If it cannot be factored or coefficients are large, use quadratic formula. x =[ -b +- sqrt(b^2-4ac)]/2a The quadratic formula is derived from completing the square. Completing the square is used to put a quadratic equation from standard form into vertex form for graphing purposes and so that "x" can be isolated. x^2 -4x -10 = 0 x^2 -4x = 10 x^2 -4x +4 = 10+4 (x-2)^2 = 14 x-2 = +-sqrt(14) x = 2 +- sqrt(14) --> end up with same answer as using quadratic formula Also benefit of doing it this way is we know the vertex of parabola vertex at (2,-14) OpenStudy (anonymous):

thanks for all the great insight. This will be very helpful for sure.

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