How do you solve using the completing the square method #x(6x - 5) = 6#?

1 Answer
May 9, 2016

Answer:

#x=+3/2" and " -2/3#

Explanation:

Multiply out the racket giving:#" "6x^2-5x=6#

Subtract 6 from both sides

#" "6x^2-5x-6=0#

Write as#" "6(x^2-5/6x)-6=0#

Take the square outside the bracket and add the correction constant k

#6(x-5/6x)^2-6+k=0#

Remove the #x# from #-5/6x#

#6(x-5/6)^2-6+k=0#

Multiply the #-5/6# by #(1/2)#

#6(x-5/(12))^2-6+k=0#

'~~~~~~~~~~~~~ Comment ~~~~~~~~~~~~~~~~~~~~~~~~~
Consider the part of #(x-5/12)(x-5/12)#

The #(-5/12)^2# is an introduced value that is not in the original equation so we remove it by subtraction. However, this introduced error is in fact #6(-5/12)^2# due to the 6 outside the brackets
#=>k=(-1)xx6xx(-5/12)^2= -1 1/24 =-25/24#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(brown)(6(x-5/(12))^2-6+k=0)" "->" "color(blue)(6(x-5/(12))^2-6-25/24=0#

#=>6(x-5/(12))^2-169/24=0#

#6(x-5/12)^2=169/24#

#(x-5/12)^2=169/144#

Taking the square root of both sides

#x-5/12=+-sqrt(169)/sqrt(144) = +-13/12#

#x=5/12+-13/12#

#x=+18/12" and " -8/12#

#x=+3/2" and " -2/3#