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

2 Answers
Mar 14, 2016

Answer:

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#

Mar 14, 2016

Answer:

#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#