How do you solve #3x^2-5x+1=0# by completing the square?

1 Answer
Oct 12, 2016

Answer:

#x=(5+sqrt13)/6 or#
#x=(5-sqrt13)/6#

Explanation:

To solve this equation we have to factorize #3x^2-5x+1#
Since we can not use any of the polynomial identities so let us
compute #color(blue)delta#

#color(blue)(delta=b^2-4ac)#
#delta=(-5)^2-4(3)(1)#
#delta=25-12=13#

The roots are :
#x_1=(-b+sqrtdelta)/(2a)=color(red)((5+sqrt13)/6)#
#x_2=(-b+sqrtdelta)/(2a)=color(red)((5-sqrt13)/6)#

Now let us solve the equation:
#3x^2-5x+1=0#
#(x-x_1)(x-x_2)=0#
#(x-color(red)((5+sqrt13)/6))(x-color(red)((5-sqrt13)/6))=0#
#x-(5+sqrt13)/6=0 rArr x=(5+sqrt13)/6 or#
#x-(5-sqrt13)/6=0rArr x=(5-sqrt13)/6#