Answer:
Hello,
3
Step-by-step explanation:
Let say x the whole number not null to find.
[tex]x+\dfrac{1}{x} =\dfrac{10}{3} \\\\\\\dfrac{x^2+1}{x} =\dfrac{10}{3} (reducing\ to\ the\ same\ denominator)\\\\3x^2+3=10x\ (cross\ products)\\\\\\3x^2-10x+3=0\ (passing\ all\ terms\ in\ the\ first\ member)\\\\\Delta=10^2-4*3*3=64=8^2\\\\x=\dfrac{10-8}{6} =\dfrac{2}{6}= \dfrac{1}{3} \ not \ a \ whole\ number\\\\x=\dfrac{10+8}{6} =3\\\\[/tex]
