How do you solve #3x^2+2=4x #?

1 Answer
Jul 29, 2016

Answer:

#=> x= 2/3+sqrt3/3i" "# and #" "x=2/3-sqrt3/3i#

#x!inRR" but "x in CC#

Explanation:

If able to do so the quickest way is to factorise.

Subtract #4x# from both sides giving:

#3x^2-4x+2=0#

Try 1
#(3x+2)(x+1) = 3x^2+3x+2x+2 color(red)(larr" Fail")#

Try2
#(3x+1)(x+2) =3x^2+6x+1x+2 color(red)(larr" Fail")#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Looks as though we need to use another approach

#color(blue)("Completing the square")#

For detailed method of approach look at
https://socratic.org/s/awA8fpNk

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

Write as: #3(x^2-4/3x)+2+k=0#

Giving:#" "3(x-4/6)^2+2-4/3=0#

#=>3(x-2/3)^2+2/3=0#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#(x-2/3)^2=-2/3xx1/3 -= -1/3#

#x-2/3=sqrt(-1/3)#

Square root of a negative number means that the curve does not cross the x-axis so there is no solution in the set of numbers called 'Real'. Written as #x!inRR#

However there is at least 1 solution in the set of number called 'Complex'. Written as #x in CC#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So we now have:

#x=2/3+-1/sqrt(3)sqrt(-1)#

#=> x= 2/3+sqrt3/3i" "# and #" "x=2/3-sqrt3/3i#

Tony B