
The sum of a number and its reciprocal is \[2\dfrac{1}{30}\]. Find the number.
Answer
626.4k+ views
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,
\[2\dfrac{1}{30}=\dfrac{61}{30}\]
Now, according to the given conditions in the question we have,
\[x+\dfrac{1}{x}=\dfrac{61}{30}\]
Taking the LCM of the left-hand side of the expression we get,
\[\begin{align}
& x+\dfrac{1}{x}=\dfrac{61}{30} \\
& \Rightarrow \dfrac{\left( x{}^\text{2}\text{ }+\text{ }1 \right)}{x}=\dfrac{61}{30} \\
\end{align}\]
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,
\[\begin{array}{*{35}{l}}
{} \\
\Rightarrow 30x{}^\text{2}\text{ }-\text{ }61x\text{ }+\text{ }30\text{ }=\text{ }0 \\
\Rightarrow 30x{}^\text{2}\text{ }-\text{ }36x\text{ }-\text{ }25x\text{ }+\text{ }30=0 \\
\Rightarrow 6x\left( 5x\text{ }-\text{ }6 \right)-5\left( 5x\text{ }-\text{ }6 \right)=0 \\
\Rightarrow \left( 6x\text{ }-\text{ }5 \right)\text{ }\left( 5x\text{ }-\text{ }6 \right)\text{ }=\text{ }0 \\
\end{array}\]
$\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.
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,
\[2\dfrac{1}{30}=\dfrac{61}{30}\]
Now, according to the given conditions in the question we have,
\[x+\dfrac{1}{x}=\dfrac{61}{30}\]
Taking the LCM of the left-hand side of the expression we get,
\[\begin{align}
& x+\dfrac{1}{x}=\dfrac{61}{30} \\
& \Rightarrow \dfrac{\left( x{}^\text{2}\text{ }+\text{ }1 \right)}{x}=\dfrac{61}{30} \\
\end{align}\]
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,
\[\begin{array}{*{35}{l}}
{} \\
\Rightarrow 30x{}^\text{2}\text{ }-\text{ }61x\text{ }+\text{ }30\text{ }=\text{ }0 \\
\Rightarrow 30x{}^\text{2}\text{ }-\text{ }36x\text{ }-\text{ }25x\text{ }+\text{ }30=0 \\
\Rightarrow 6x\left( 5x\text{ }-\text{ }6 \right)-5\left( 5x\text{ }-\text{ }6 \right)=0 \\
\Rightarrow \left( 6x\text{ }-\text{ }5 \right)\text{ }\left( 5x\text{ }-\text{ }6 \right)\text{ }=\text{ }0 \\
\end{array}\]
$\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.
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