Chapter 7.4, Problem 3P

To determine

To find: The number whose sum of reciprocal of two consecutive odd integers is $\frac{8}{15}$ .

Answer to Problem 3P

The number is $3$ and $\frac{-5}{4}$ .

Explanation of Solution

Given information:

The sum of reciprocal of two consecutive odd integers is $\frac{8}{15}$ .

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 odd integers is $x$ and $x+2$ .

The sum of reciprocal is $\frac{1}{x}+\frac{1}{x+2}=\frac{8}{15}\mathrm{..........}\left(i\right)$ .

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

Therefore $\frac{1}{x}+\frac{1}{x+2}=\frac{8}{15}$

  $\Rightarrow \frac{x+2+x}{x\left(x+2\right)}=\frac{8}{15}\Rightarrow \frac{2x+2}{x^2+2x}=\frac{8}{15}$ .

We cross multiply the term $\frac{2x+2}{x^2+2x}=\frac{8}{15}$ .

Therefore we have $15*\left(2x+2\right)=8*\left(x^2+2x\right)\Rightarrow 30x+30=8x^2+16x\Rightarrow 8x^2-14x-30$ .

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

Where $a=8,b=-14,c=-15$ .

Therefore

  $\begin{array}{l}\frac{-\left(-14\right)\pm \sqrt{{\left(-14\right)}^2+4*8*30}}{2*8}\Rightarrow \frac{14\pm \sqrt{196+960}}{16}\Rightarrow \frac{14\pm \sqrt{1156}}{16}\\ \Rightarrow \frac{14\pm 34}{16}\Rightarrow \frac{14+34}{16},\frac{14-34}{16}\Rightarrow \frac{48}{16},\frac{-20}{16}\Rightarrow 3,\frac{-5}{4}\end{array}$

Therefore the number is $3$ and $\frac{-5}{4}$ whose sum of reciprocal odd integers is $\frac{8}{15}$ .