How do you solve #\frac { 1} { 2} x ^ { 2} - 3x + 1= 0#?

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Answer 1 · 2017-04-13T21:12:41

#x = 3+-sqrt7#

#x = 5.646 " or " x = 0.354#

#1/2x^2-3x+1=0#

Multiply through by 2 to get rid of the fraction:

#x^2 -6x+2=0#

There are no factors of #2# which add up to #6#, so this expression does not factorise.

We can complete the square easily because #b# is even.

#x^2 -6x color(red)(+9) =-2color(red)(+9)" "larrcolor(red)(((-6)/2)^2 =9)#
#(x-3)^2= 7#

#x-3 = +-sqrt7#

#x = 3+-sqrt7#