#2x + 2/x = 3#
First we have to find the value of #x# before looking for what is asked..
#(2x)/1 + 2/x = 3/1#
Multiply through with the LCM which is #x# in this case
#x((2x)/1) + cancelx(2/cancelx) = x(3/1)#
#x(2x) + 2 = x(3)#
#2x^2 + 2 = 3x#
#2x^2 - 3x + 2 = 0 -> "Quadratic Equation"#
Well in solving a Quadratic Equation you have to find the sum and products of the roots..
In this case the possible roots are #4# and #1#
In the above equation, you multiply #2# attached to the square of #x# with the constant..
Hence #2 xx 2 = 4#
Therefore, we are going to look for a possible value that can multiply each other to get #+4# and can either subtract or add each other to get #-3#
Hence #4 and 1# are the roots..
Given that #rArr 4 xx 1 = 4 and -4 + 1 = -3#
Now Solving...
#2x^2 color(red)(- 4)x color(red)(+1) x + 1 = 0#
#(2x^2 - 4x) (- x + 2) = 0#
Factorizing..
#2x(x - 2) -1(x - 2) = 0#
#(x - 2) (2x - 1) = 0#
#x - 2 = 0#
#x = 2#
#or#
#2x - 1 = 0#
#2x = 1#
#x= 1/2#
Hence we should use #x = 2# since it's positive
Now..
Find, #x^3 + 1/x + 2#
Substitute the value of #x# into the equation..
#x^3 + 1/x + 2 -> "Equation"#
#color(red)2^3 + 1/color(red)2 + 2#
#8 + 1/2 +2#
#8 + 2 + 1/2#
#10 + 1/2#
#10 1/2#
