Question

The sum of the squares two consecutive intergers is 45 . What are the intergers

Answer 1

Let's denote the two consecutive integers as \( n \) and \( n + 1 \). The equation based on the problem's statement is:

\[
n^2 + (n + 1)^2 = 45
\]

Now, expand the equation:

\[
n^2 + (n^2 + 2n + 1) = 45
\]

Combine like terms:

\[
2n^2 + 2n + 1 = 45
\]

Subtract 45 from both sides:

\[
2n^2 + 2n + 1 - 45 = 0
\]

This simplifies to:

\[
2n^2 + 2n - 44 = 0
\]

Now, divide the entire equation by 2:

\[
n^2 + n - 22 = 0
\]

Next, we can factor this quadratic equation (or use the quadratic formula). We need two numbers that multiply to \( -22 \) and add to \( 1 \). The numbers \( 11 \) and \( -2 \) fit, so we can factor it as:

\[
(n + 11)(n - 2) = 0
\]

Setting each factor equal to zero gives us the possible solutions for \( n \):

\[
n + 11 = 0 \quad \Rightarrow \quad n = -11
\]

\[
n - 2 = 0 \quad \Rightarrow \quad n = 2
\]

Thus, the pairs of consecutive integers are:

1. For \( n = -11 \): The integers are \( -11 \) and \( -10 \).
2. For \( n = 2 \): The integers are \( 2 \) and \( 3 \).

Therefore, the consecutive integers that satisfy the equation are \( (-11, -10) \) and \( (2, 3) \).