Question

How do you solve `4x^2-8x + 1 = 0`?

Answer 1

`x_1=1+sqrt(3)/2 or x_2=1-sqrt(3)/2`

Explanation:

`4x^2-8x + 1 = 0`

`4x^2-8x + 1 = 0|:4`
`x^2-2x+1-1+1/4=0`
`(x-1)^2-1+1/4=0`
`(x-1)^2-3/4=0|+3/4`
`(x-1)^2=3/4|sqrt()`
`x-1=+-sqrt(3)/2|+1`
`x=1+-sqrt(3)/2`

Answer 2

`x=1+-sqrt(3)/2` Exact values

`x~~0.13` to 2 decimal places
`x~~1.87` to 2 decimal places

Explanation:

Completing the square.

Given: `y=0=4x^2-8x+1`

Write as:
`y=0=4(x^2-8/4x)+1`

We now change this into the format of completing the square but in doing so we introduce a value that is not in the original equation. We 'get rid' of this by introducing a value that changes it into 0. Let this as yet unknown value be `k`

`0=4(x^2color(white)("d")ubrace(-8/4)color(white)("d")x)+1`
`color(white)("dddddddd.d")darr`
`color(white)("ddddddd")"Halve this"`
`color(white)("dddddddd.d")darr`
`0=color(white)("d")4(xcolor(white)("d.")obrace(-1)color(white)("d"))^2+k+1`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Set `4(-1)^2+k=0 color(white)("dddd")=>color(white)("dddd")k=-4`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`0=4(x-1)^2-4+1`

`0=4(x-1)^2-3`

`(x-1)^2=3/4`

`x-1=+-sqrt(3)/2`

`x=1+-sqrt(3)/2` Exact values

`x~~0.13397...`
`x~~1.86602...`

`x~~0.13` to 2 decimal places
`x~~1.87` to 2 decimal places

Answer 3

`x=1+sqrt 3/2` or `x=1-sqrt 3/2`

Explanation:

`4x^2-8x+1=0`

Using the quadratic formula

`:.ax^2+bx+c=0`

`:.a=4,b=-8,c=1`

`:.x=(-b+-sqrt(b^2-4ac))/(2a)`

`:.x=(-(-8)+-sqrt((-8)^2-4(4)(1)))/(2(4))`

`:.x=(8+-sqrt 48)/8`

`:.x=(sqrt(4*4*3))/8`

`:.x=(8+-4 sqrt 3)/8`

`:.x=cancel 8^1/cancel 8^1+-(cancel4^1 sqrt 3)/cancel 8^2`

`:.x=1+-sqrt 3/2`

`:.x=1+sqrt 3/2 or x=1-sqrt 3/2`