Chapter 7.4, Problem 4P

To determine

To find: The number whose sum of reciprocal of two consecutive even integers is $\frac{11}{60}$ .

Answer to Problem 4P

The number is $10$ and $\frac{-12}{11}$ .

Explanation of Solution

Given information:

The sum of reciprocal of two consecutive even integers is $\frac{11}{60}$ .

Formula used:

We have to equate both left and right hand side.

And then we will use quadratic equation $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Calculation:

Consider that the two consecutive even integers is $x$ and $x+2$ .

The sum of reciprocal is $\frac{1}{x}+\frac{1}{x+2}=\frac{11}{60}\mathrm{..........}\left(i\right)$ .

Recall that we have to equate both left and right hand side of the equation $\left(i\right)$ .

Therefore $\frac{1}{x}+\frac{1}{x+2}=\frac{11}{60}$ $\Rightarrow \frac{x+2+x}{x\left(x+2\right)}=\frac{11}{60}\Rightarrow \frac{2x+2}{x^2+2x}=\frac{11}{60}$ .

We cross multiply the term $\frac{2x+2}{x^2+2x}=\frac{11}{60}$ .

Therefore we have

  $60*\left(2x+2\right)=11*\left(x^2+2x\right)\Rightarrow 120x+120=11x^2+22x\Rightarrow 11x^2-98x-120=0$

Recall that we will use quadratic equation $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Where $a=11,b=-98,c=-120$ .

Therefore

$\begin{array}{l}\frac{-\left(-98\right)\pm \sqrt{{\left(-98\right)}^2+4*11*120}}{2*11}\Rightarrow \frac{98\pm \sqrt{9604+5280}}{22}\Rightarrow \frac{98\pm \sqrt{14884}}{22}\\ \Rightarrow \frac{98\pm 122}{22}\Rightarrow \frac{98+122}{22},\frac{98-122}{22}\Rightarrow \frac{220}{22},\frac{-24}{22}\Rightarrow 10,\frac{-12}{11}\end{array}$

Therefore the number is $10$ and $\frac{-12}{11}$ whose sum of reciprocal odd integers is $\frac{11}{60}$ .