Question

A natural number when increased by 84 equals 460 times its reciprocal. Find the number.

Answer 1

Hint: Before solving this type of problem we have to get the knowledge of the number system and the natural number, now we have to assume the number as some variable and add 84 in it. Then we have to equate the new number with 160 times of the assumed variable reciprocal.

Complete step-by-step answer:
Given:
An unknown natural number increased by 84 equals to 160 times of its reciprocal.
We have to assume the unknown natural number be x.
Now we have to add 84 to the unknown natural number then the new will be, (x + 84) …… (1)
Then the reciprocal of the unknown natural number will be, $ \dfrac{1}{x} $.
As it is given that the reciprocal of unknown natural numbers is 160 times. So, (160/x) …… (2)
Now we have to equate the expression (1) and (2) as follows:
$\begin{align}
\Rightarrow & x+84=\dfrac{160}{x} \\
\Rightarrow & {{x}^{2}}+84x=160 \\
\Rightarrow & {{x}^{2}}+84x-160=0......\ \left( 3 \right)
\end{align}$
Find the value of x using the quadratic formula given below.
${{x}_{1,2}}=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
From the expression (3) we have to find the variables as follows: a = 1, b = 84 and c = -160.
We have to substitute the values of a, b, and c in the quadratic formula to find the value of x.
$\begin{align}
\Rightarrow & {{x}_{1,2}}=\dfrac{-84\pm \sqrt{{{\left( 84 \right)}^{2}}-4\left( 1 \right)\left( -160 \right)}}{2\times 1} \\
 & =\dfrac{-84\pm 4\sqrt{481}}{2} \\
 & =-42\pm 2\sqrt{481} \\
 & {{x}_{1}}=-42+2\sqrt{481} \\
 & {{x}_{2}}=-42-2\sqrt{481}
\end{align}$
By solving the above expression, we get the required unknown natural numbers as follows: $-42 + 2\sqrt{481}$ and -42 - $2\sqrt{481}$.

Note: We have to be careful during the formulation of the equation and also during the solution of quadratic formula for x. We have to be careful during solving the quadratic formula for roots. Remember that the quadratic equation has two roots so consider both.