#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#