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

3 Answers
May 11, 2018

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

May 11, 2018

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

May 11, 2018

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