Question

Find a number such that the sum and three times its reciprocal is 91/20.

Answer 1

For this exercise you need to remember the following:

1. The sum is the result of an Addition.

2. By definition, given:

$\frac{a}{b}$

Its reciprocal is:

$=\frac{b}{a}$

3. The word "times" indicates a Multiplication.

In this case, let be "n" the number mentioned in the exercise. Its reciprocal is:

$\frac{1}{n}$

Based on the information given in the exercise, you can set up the following equation:

$n+3(\frac{1}{n})=\frac{91}{20}$

Now you must solve for "n":

1. Simplify:

$\begin{gathered} n+\frac{3}{n}=\frac{91}{20} \\ \\ \frac{(n)(n)+3}{n}=\frac{91}{20} \\ \\ \frac{n^2+3}{n}=\frac{91}{20} \\ \\ n^2+3=\frac{91}{20}n \end{gathered}$

2. Make the equation equal to zero:

$n^2-\frac{91}{20}n+3=0$

3. Apply the Quadratic formula. This is:

$n=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

In this case:

$\begin{gathered} a=1 \\ b=-\frac{91}{20}_{} \\ \\ c=3 \end{gathered}$

Substituting values and evaluating, you get:

$\begin{gathered} n=\frac{-(-\frac{91}{20})\pm\sqrt[]{(-\frac{91}{20})^2-2(1)(3)}}{2\cdot1} \\ \\ n_1=\frac{15}{4} \\ \\ n_2=\frac{4}{5} \end{gathered}$

The answer

There are two numbers:

$\begin{gathered} n_1=\frac{15}{4} \\ \\ n_2=\frac{4}{5} \end{gathered}$