How do you solve #4x^2 - 12x - 3 = 0# by completing the square?

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Answer 1 · 2017-06-12T19:06:29

#x=3/2+-sqrt(3) #

#x~~-0.23 and x~~+3.23 #

Given:#" "0=4x^2-12x-3#

For a complete explanation of the method see:
https://socratic.org/s/aFvmbGAs

It uses different numbers.

Set #y=0=4x^2-12x-3#

#y=0=4(x-12/(4xx2))^2+k-3#

#y=0=4(x-3/2)^2+k-3#
.........................................................................

Set #" "4(-3/2)^2+k=0#

#4xx9/4+k=0" "=>" "k =-9#
.......................................................................

#y=0=4(x-3/2)^2-12#

#+12/4=(x-3/2)^2#

#x-3/2=+-sqrt(3)#

Exact answer is:
#x=3/2+-sqrt(3) #

Approximate answer is:
#x=-0.232050.. and x=+3.232-50....#

Tony B