Question & Answer

The sum of a number and its reciprocal is \[2\dfrac{1}{30}\]. Find the number.

ANSWER Verified Verified
Hint: To solve such types of questions we first of all assume the number as some variable x then the reciprocal of the number would become \[\dfrac{1}{x}\]. Now use the given conditions in the question to formula a polynomial in terms of the variable x and then calculate the value of the number.

Complete step-by-step answer:
Let the one number be x
So, its reciprocal will be\[\dfrac{1}{x}\].
Given that the sum of a number and its reciprocal is \[2\dfrac{1}{30}\].
Converting \[2\dfrac{1}{30}\] into fraction form we have,
Now, according to the given conditions in the question we have,
Taking the LCM of the left-hand side of the expression we get,
  & x+\dfrac{1}{x}=\dfrac{61}{30} \\
 & \Rightarrow \dfrac{\left( x{}^\text{2}\text{ }+\text{ }1 \right)}{x}=\dfrac{61}{30} \\
After cross multiplying the above obtained expression, we get
\[\Rightarrow 30x{}^\text{2}\text{ }+\text{ }30=61x\]
Making necessary arrangements we get,
 \[\Rightarrow 30x{}^\text{2}\text{ }-\text{ }61x\text{ }+\text{ }30\text{ }=\text{ }0\]
Now we have obtained a quadratic expression. Solving it by splitting middle terms we get,
   {} \\
   \Rightarrow 30x{}^\text{2}\text{ }-\text{ }61x\text{ }+\text{ }30\text{ }=\text{ }0 \\
   \Rightarrow 30x{}^\text{2}\text{ }-\text{ }36x\text{ }-\text{ }25x\text{ }+\text{ }30=0 \\
   \Rightarrow 6x\left( 5x\text{ }-\text{ }6 \right)-5\left( 5x\text{ }-\text{ }6 \right)=0 \\
   \Rightarrow \left( 6x\text{ }-\text{ }5 \right)\text{ }\left( 5x\text{ }-\text{ }6 \right)\text{ }=\text{ }0 \\
  $\Rightarrow 6x = 5 $ and $5x = 6$
  $ \Rightarrow x={\dfrac{5}{6}}, {\dfrac{6}{5}}$

Therefore the number is either \[\dfrac{5}{6}\] or \[\dfrac{6}{5}\]

Note: Students can assume both the numbers as different variables which would become complex and while making an equation of the given condition it will be a polynomial or equations in two variables which would be difficult to solve. Therefore, always assume a single variable for such types of equations and try to generalise the second variable in terms of the first one.