Hiiii ppls.

I plugged into Wolfram Alpha, and got 19/3.

Use the quadratic formula to find a, b:

x = 1/10 (11 - sqrt(41)) = a
x = 1/10 (11 + sqrt(41))= b

1 / [1/10 (11 + sqrt(41))] + 1 / [1/10 (11 - sqrt(41))] 

=11 / 4

\(If \qquad 5x^2 - 11x + 4 = 0\\ \text{and a and b are the solutions then}\\ a=\frac{11+\sqrt{41}}{10}\qquad and \qquad b=\frac{11-\sqrt{41}}{10}\\~\\ \frac{1}{a}+\frac{1}{b}=\frac{10}{11+\sqrt{41}}+\frac{10}{11-\sqrt{41}}\\~\\ \frac{1}{a}+\frac{1}{b}=\frac{10(11-\sqrt{41})+10(11+\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\ \frac{1}{a}+\frac{1}{b}=\frac{(110-10\sqrt{41})+(110+10\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\ \frac{1}{a}+\frac{1}{b}=\frac{(110-10\sqrt{41})+(110+10\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\ \)

I think you can finish it from there.

LaTex:

If \qquad 5x^2 - 11x + 4 = 0\\ \text{and a and b are the solutions then}\\

a=\frac{11+\sqrt{41}}{10}\qquad or \qquad b=\frac{11-\sqrt{41}}{10}\\~\\

\frac{1}{a}+\frac{1}{b}=\frac{10}{11+\sqrt{41}}+\frac{10}{11-\sqrt{41}}\\~\\

\frac{1}{a}+\frac{1}{b}=\frac{10(11-\sqrt{41})+10(11+\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\

\frac{1}{a}+\frac{1}{b}=\frac{(110-10\sqrt{41})+(110+10\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\

\frac{1}{a}+\frac{1}{b}=\frac{(110-10\sqrt{41})+(110+10\sqrt{41})}{(11+\sqrt{41})(11-\sqrt{41})}\\~\\

If the roots of the equation are a and b, then

a + b = 11/5, and ab = 4/5.

So,

\(\displaystyle \frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{11}{5}.\frac{5}{4}=\frac{11}{4}.\)