```Question 18721
SEE THIS EXAMPLE WHICH IS MORE COMPLEX AND EXPLANATION  TO SOLVE YOUR PROBLEM
x - 2/x + 3 is less than 1
let y=(x - 2/x + 3)<1
x - 2/x + 3-1<0 ....or....x - 2/x + 2<0
(x^2-2+2x)/x<0...or ...(x^2+2x-2)/x<0..
now a fraction will be -ve (<0)if n.r and d.r are of different signs..let us take the 2 cases
dr=x is +ve...then nr should be -ve
dr=x is -ve...then nr should be +ve.
now solve the nr  using quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (-2 +- sqrt( 2^2-4*1*(-2)))/(2*1) }}}
hence x=(-1+sqrt3)/2...and....(-1-sqrt3)/2..for conveinience if we call these 2 values as  p(approximately=-1.37)and q(approximately=0.37),we find that if x lies between p  and q (-1.37 and 0.37 )nr is -ve and when x is less than
p(-1.37) or greater than q(0.37) ,nr is +ve.
now we have to combine this with the above assumption on dr
dr=x is +ve...then nr should be -ve..so x should be between 0 and (-1+sqrt3)/2(not -1.37 to zero as x is already taken as positive)
dr=x is -ve...then nr should be +ve.so x should be less than ....(-1-sqrt3)/2..
(not 0.37 to zero as x is already taken as negative)

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