Question

Two positive numbers and their average have a sum of 63. One of the two numbers is the square of the other. What is the smaller of the two numbers?

Answer 1

Let the two positive numbers be \( x \) and \( y \), where we can assume that \( y = x^2 \) (since one number is the square of the other).

The average of the two numbers is given by:

\[
\text{Average} = \frac{x + y}{2} = \frac{x + x^2}{2}
\]

According to the problem, the sum of the two numbers \( x + y \) and their average is equal to 63:

\[
x + y + \frac{x + y}{2} = 63
\]

We can substitute \( y = x^2 \) into the equation:

\[
x + x^2 + \frac{x + x^2}{2} = 63
\]

Now, let's express the left side. First, simplify \( x + x^2 \):

\[
x + x^2 = x + x^2
\]

Now, consider the average:

\[
\frac{x + x^2}{2}
\]

Thus, we can rewrite the equation as:

\[
x + x^2 + \frac{x + x^2}{2} = 63
\]

This leads us to a common denominator of 2:

\[
\frac{2(x + x^2) + (x + x^2)}{2} = 63
\]

Combining the numerators:

\[
\frac{3(x + x^2)}{2} = 63
\]

To eliminate the fraction, multiply both sides by 2:

\[
3(x + x^2) = 126
\]

Next, divide both sides by 3:

\[
x + x^2 = 42
\]

Now, rearranging gives us a quadratic equation:

\[
x^2 + x - 42 = 0
\]

We can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 1, c = -42 \):

\[
b^2 - 4ac = 1^2 - 4(1)(-42) = 1 + 168 = 169
\]

Thus, we have:

\[
x = \frac{-1 \pm \sqrt{169}}{2}
\]

Calculating the square root, we find:

\[
\sqrt{169} = 13
\]

Substituting back, we have:

\[
x = \frac{-1 \pm 13}{2}
\]

This results in two potential solutions:

1. \( x = \frac{12}{2} = 6 \)
2. \( x = \frac{-14}{2} = -7 \) (not a valid solution since \( x \) must be positive)

Therefore, we find:

\[
x = 6
\]

Since \( y = x^2 \):

\[
y = 6^2 = 36
\]

The smaller of the two numbers \( x \) and \( y \) is clearly:

\[
\boxed{6}
\]

To verify, we check the initial condition that the sum of the two numbers and their average equals 63:

\[
6 + 36 + \frac{6 + 36}{2} = 6 + 36 + 21 = 63
\]

Thus, the calculations hold true, and the smaller number is \( \boxed{6} \).