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Mathematics 8 Online
OpenStudy (anonymous):

what is the checking for completing the square ?? hao it is ??? please help me ! :((

OpenStudy (anonymous):

this is the problem .. what are the zeros of f(x)= 2x(squared) - 5x + 1

OpenStudy (anonymous):

its the process of rearranging a trinomial so that it contains a perfect square for example x^2 + 4x - 1 = (x + 2)^2 -2^2 - 1 = (x + 2)^2 - 5

OpenStudy (anonymous):

i already get the correct answer my problem is the checking . :((

OpenStudy (anonymous):

Substitute your answers back into the original equation and see if it is 0

OpenStudy (anonymous):

but how ??

OpenStudy (anonymous):

help me please ?? :(

OpenStudy (akshay_budhkar):

for checking you just substitute the values you got.. what did you get as your answer?

OpenStudy (akshay_budhkar):

tine??????

OpenStudy (anonymous):

the first one is the square root of 17/16 . that is from the first equation.. the second one is the value of x . it is x=5(square root)17/4.... and the third one is x= -1/5 ... :)

OpenStudy (anonymous):

Equation is a quadratic, there are 2 solutions.

OpenStudy (anonymous):

sorry . it is x=5(square root)17/4 and x= -1/5 . :)

OpenStudy (anonymous):

can you help me for this ??

OpenStudy (anonymous):

Your original equation is 2x(squared) - 5x + 1 Let's take x= -1/5 -> 2(-1/5)^2 -5(-1/5) + 1 = 2/25 +1 +1 = 2 and 2/25 not 0 so this solution is wrong.

OpenStudy (anonymous):

Completed square and solutions are there.

OpenStudy (anonymous):

aww . how about if x=5(square root)17/4 is x=-5(square root)17/4 ???

OpenStudy (anonymous):

The solutions at the link I gave u are 5/4 plus or minus (square root)17/4

OpenStudy (anonymous):

how about if x=5(square root)17/4 is x=-5(square root)17/4 ??? can you solve it again ??? i think you can get the zero now. :(

OpenStudy (anonymous):

please ?? if you dont mind. :)

OpenStudy (anonymous):

The solutions are at the link I gave u and repeated above. There is no point in checking wrong solutions, it is a waste of time, mine and yours.

OpenStudy (anonymous):

ok ! thanks. :)

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