Understanding the Problem: Sum, Product, and Reciprocals

The problem asks us to find the sum of the reciprocals of two positive numbers given their sum and the square root of their product. Let the two positive numbers be \(a\) and \(b\).

Given Information:

• The sum of the two positive numbers is 65. Mathematically, this can be written as:

  \(a + b = 65\)

• The square root of their product is 26. Mathematically, this is:

  \(\sqrt{ab} = 26\)

Goal:

We need to find the sum of their reciprocals. The reciprocal of a number \(x\) is \(\frac{1}{x}\). So, the sum of the reciprocals of \(a\) and \(b\) is \(\frac{1}{a} + \frac{1}{b}\).

Solving the Problem Step-by-Step

Step 1: Find the product of the two numbers

We are given that the square root of the product is 26. To find the product \(ab\), we need to square both sides of the equation \(\sqrt{ab} = 26\).

\((\sqrt{ab})^2 = 26^2\)

\(ab = 26 \times 26\)

Calculating the square of 26:

\(26 \times 26 = 676\)

So, the product of the two numbers is:

\(ab = 676\)

Step 2: Express the sum of reciprocals in terms of the sum and product

The expression we need to find is \(\frac{1}{a} + \frac{1}{b}\). To combine these fractions, we find a common denominator, which is \(ab\).

\(\frac{1}{a} + \frac{1}{b} = \frac{1 \times b}{a \times b} + \frac{1 \times a}{b \times a}\)

\(\frac{1}{a} + \frac{1}{b} = \frac{b}{ab} + \frac{a}{ab}\)

\(\frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}\)

Step 3: Substitute the known values and simplify

We know the sum \(a + b = 65\) (given) and the product \(ab = 676\) (calculated in Step 1). Substitute these values into the expression for the sum of reciprocals:

\(\frac{a + b}{ab} = \frac{65}{676}\)

Now, we need to simplify the fraction \(\frac{65}{676}\).

We can notice that 65 is \(5 \times 13\). Let's check if 676 is divisible by 13.

\(676 \div 13\)

Using division:

\(676 = 13 \times 52\)

So, the fraction becomes:

\(\frac{65}{676} = \frac{5 \times 13}{52 \times 13}\)

Cancel out the common factor of 13:

\(\frac{5}{52}\)

Thus, the sum of the reciprocals of the two numbers is \(\frac{5}{52}\).

Summary of Calculations

Revision Table: Key Concepts Reviewed

Additional Information: Relation to Quadratic Equations

This problem can also be related to finding the numbers themselves, although it wasn't necessary to solve the sum of reciprocals. If we know the sum \(a+b\) and the product \(ab\), \(a\) and \(b\) are the roots of the quadratic equation:

\(x^2 - (a+b)x + ab = 0\)

Using the given values:

\(x^2 - 65x + 676 = 0\)

We could solve this quadratic equation to find the values of \(a\) and \(b\). For example, using the quadratic formula \(\frac{-B \pm \sqrt{B^2-4AC}}{2A}\) for \(Ax^2+Bx+C=0\):

\(x = \frac{-(-65) \pm \sqrt{(-65)^2 - 4(1)(676)}}{2(1)}\)

\(x = \frac{65 \pm \sqrt{4225 - 2704}}{2}\)

\(x = \frac{65 \pm \sqrt{1521}}{2}\)

\(x = \frac{65 \pm 39}{2}\)

The two numbers are:

• \(x_1 = \frac{65 + 39}{2} = \frac{104}{2} = 52\)
• \(x_2 = \frac{65 - 39}{2} = \frac{26}{2} = 13\)

So the two positive numbers are 13 and 52. Let's verify:

• Sum: \(13 + 52 = 65\) (Correct)
• Product: \(13 \times 52 = 676\) (Correct)
• Square root of product: \(\sqrt{676} = 26\) (Correct)

Now, let's find the sum of their reciprocals using these numbers:

\(\frac{1}{13} + \frac{1}{52}\)

The common denominator is 52.

\(\frac{1 \times 4}{13 \times 4} + \frac{1}{52} = \frac{4}{52} + \frac{1}{52}\)

\(\frac{4 + 1}{52} = \frac{5}{52}\)

This confirms the result obtained earlier by just using the sum and product properties, which was a more direct method for this specific question.