How do you solve #3x^2 - 5x+4=0# by completing the square?

2 Answers
May 10, 2018

Answer:

#3(x-5/6)^2+23/12#

Explanation:

#3x^2-5x+4#
#3(x^2-5/3x+4/3)#
#3((x-5/6)^2-(5/6)^2+4/3)#
#3(x-5/6)^2-3(5/6)^2+4#
#3(x-5/6)^2-25/12+4#
#3(x-5/6)^2+23/12#

I double checked and worked backwards and gives you the original function. I knew I was working with fractions the minute I saw the #3x^2# which I didn't mind, as long as you prioritise the #3x^2# and turn that into #x^2# by itself so that completing the square would be easier.

May 10, 2018

Answer:

#x=5/6+-sqrt(23)/6 i#

Explanation:

Write as:

#3(x^2-5/3x)+4=0" "................Equation(1)#

The next step introduces a value that is not in the original equation. We get rid of it by introducing the the as yet unknown value #k#. We will use this to turn the introduced value into 0

#color(green)(color(white)("dd.d")3(x^2-5/3x)+4=0)#
#color(white)("ddddd.dddd")darr#

#"halve the " -5/3#

#color(white)("ddddd.dddd")darr#
#color(green)(color(white)("ddd.d")3(x-5/6)^2+k+4=0)" ".............Equation(2)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The introduced value is #3(-5/6)^2# so:

Set: #3(-5/6)^2+k=0# giving

#3(+25/36)+k=0#

#25/12+k=0color(white)("d")=>color(white)("d")k=-25/12" "......Equation(3)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute #Eqn(3)# into #Eqn(2)#

#3(x-5/6)^2-25/12+4=0#

#3(x-5/6)^2+23/12=0#

#(x-5/6)^2=-23/36#

#x-5/6=+-sqrt(-23/36)#

#x=5/6+-sqrt(23)/6 i#

Tony B