Question

How do you solve using the completing the square method `4t^2 = 8t - 1`?

Answer 1

See details below to obtain
`color(white)("XXX")t=1+-sqrt(3)/2`

Explanation:

Given
`color(white)("XXX")4t^2=8t-1`

Get all terms involving the variable on the left side
`color(white)("XXX")4t^2-8t=-1`

Divide both sides by `4`
`color(white)("XXX")t^2-2t= -1/4`

If `t^2-2t` are the first two terms of an expanded squared binomial,
then the third term must be `1`.
Complete the square by adding `1` to both sides:
`color(white)("XXX")t^2-2t+1 = 3/4`

Re-write the left side as a squared binomial
`color(white)("XXX")(t-1)^2=3/4`

Taking the square roots:
`color(white)("XXX")t-1=+-sqrt(3)/2`

Add `1` to both sides:
`color(white)("XXX")t=1-sqrt(3)/2 and t=1+sqrt(3)/2`

Answer 2

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

Explanation:

Move all values with `t` to the left hand side.

`4t^2-8t=-1`

Divide everything by the initial coefficient `4` to make `t^2` have a coefficient of 1

`t^2-2t=-1/4`. Note `b=-2`

Add the square of half the coefficient `b` to both sides

`t^2-2t+(-2/2)^2=-1/4+(-2/2)^2`

Rewrite in perfect square form and simplify the right hand side

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

Take `+-` the square root of both sides

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

Add one to both sides

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