How do you solve for x in #1/x + x = 4#?

1 Answer

Refer to explanation

Explanation:

First the equation holds for #R-{0}#.Multiply with x both parts of equation you get

#x*(1/x+x)=4*x=>1+x^2=4x=>x^2-4x+1=0=> #

This is a trinomial with roots given by

#x_1,_2=(-b+-sqrt(b^2-4ac))/(2a)#

where #a=1,b=-4,c=1#

So the roots are
#x_1=2-sqrt3# and #x_2=2+sqrt3#

Both roots are acceptable since they belong to #R-{0}#