Question

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

Answer 1

#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`