Answer :
Answer:
3/7
Explanation:
Let the fraction be x/y.
Since the denominator is one more than twice the numerator, we can write:
y = 2x + 1
The sum of the fraction and its reciprocal is:
x/y + y/x = 32x
Simplifying the equation, we get:
(x^2 + y^2) / xy = 32x
Substituting y = 2x + 1, we get:
(x^2 + (2x + 1)^2) / x(2x + 1) = 32x
Solving the equation, we get:
x = 3
y = 2(3) + 1 = 7
So, the fraction is 3/7.
Answer:
Let the fraction be represented as \( \frac{n}{d} \), where \( n \) is the numerator and \( d \) is the denominator.
From the problem statement:
1. The denominator \( d \) is one more than twice the numerator \( n \):
\[ d = 2n + 1 \]
2. The sum of the fraction and its reciprocal is 32:
\[ \frac{n}{d} + \frac{d}{n} = 32 \]
Now substitute \( d = 2n + 1 \) into the second equation:
\[ \frac{n}{2n+1} + \frac{2n+1}{n} = 32 \]
To eliminate the fractions, find a common denominator which is \( n(2n+1) \):
\[ \frac{n^2 + (2n+1)^2}{n(2n+1)} = 32 \]
Expand \( (2n+1)^2 \):
\[ (2n+1)^2 = 4n^2 + 4n + 1 \]
So the equation becomes:
\[ \frac{n^2 + 4n^2 + 4n + 1}{n(2n+1)} = 32 \]
\[ \frac{5n^2 + 4n + 1}{n(2n+1)} = 32 \]
Cross-multiply to eliminate the fraction:
\[ 5n^2 + 4n + 1 = 32n(2n+1) \]
\[ 5n^2 + 4n + 1 = 64n^2 + 32n \]
Move all terms to one side to form a quadratic equation:
\[ 0 = 64n^2 + 32n - 5n^2 - 4n - 1 \]
\[ 0 = 59n^2 + 28n - 1 \]
Now, solve this quadratic equation using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 59 \), \( b = 28 \), and \( c = -1 \):
\[ n = \frac{-28 \pm \sqrt{28^2 - 4 \cdot 59 \cdot (-1)}}{2 \cdot 59} \]
\[ n = \frac{-28 \pm \sqrt{784 + 236}}{118} \]
\[ n = \frac{-28 \pm \sqrt{1020}}{118} \]
\[ n = \frac{-28 \pm 2\sqrt{255}}{118} \]
Simplify the expression for \( n \).
Once \( n \) is found, substitute back into \( d = 2n + 1 \) to find \( d \).
Finally, the fraction is \( \frac{n}{d} \).