How would you complete the square for this problem:
f(x)=x^2-x+(5/4)
There doesn't seem to be a way to do it:
(x^2-x+1-1)+(5/4)
(x^2-x+1)+(1/4)
So you have to factor the trinomial:
It doesn't work if you do (x-1)(x-1) and it doesn't work if you do (x-(1/2))(x-(1/2))
f(x)=x^2-x+(5/4)
There doesn't seem to be a way to do it:
(x^2-x+1-1)+(5/4)
(x^2-x+1)+(1/4)
So you have to factor the trinomial:
It doesn't work if you do (x-1)(x-1) and it doesn't work if you do (x-(1/2))(x-(1/2))
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Your mistake was at
(x^2-x+1-1)+(5/4)
It should have been
(x^2 - x + 1/4 - 1/4) + 5/4
(x - 1/2)^2 + 1
(x^2-x+1-1)+(5/4)
It should have been
(x^2 - x + 1/4 - 1/4) + 5/4
(x - 1/2)^2 + 1
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x^2 -x +(5/4) you simply half the coefficent of the "x" term then
subtract its sqaure:
(x -1/2)^2 - 1/4 + (5/4) = (x - 1/2)^2 + 1
note since f(x) > 0 for all x there is no solt. to f(x) = 0 here!
subtract its sqaure:
(x -1/2)^2 - 1/4 + (5/4) = (x - 1/2)^2 + 1
note since f(x) > 0 for all x there is no solt. to f(x) = 0 here!
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Yes you can.
example:
x^2 - 4x + 7
(x^2 - 4x + 4) + 7-4
(x-2)^2 + 3
example:
x^2 - 4x + 7
(x^2 - 4x + 4) + 7-4
(x-2)^2 + 3