Let the two consecutive integers be \( n \) and \( n + 1 \). The sum of the squares of these integers can be expressed as:
\[ n^2 + (n + 1)^2 = 145 \]
Now, expanding the equation:
\[ n^2 + (n^2 + 2n + 1) = 145 \]
Simplifying this gives:
\[ 2n^2 + 2n + 1 = 145 \]
Subtracting 145 from both sides:
\[ 2n^2 + 2n + 1 - 145 = 0 \]
This simplifies to:
\[ 2n^2 + 2n - 144 = 0 \]
Dividing the entire equation by 2:
\[ n^2 + n - 72 = 0 \]
Next, we need to factor the quadratic equation. We are looking for two numbers that multiply to \(-72\) and add to \(1\). The factors are \(9\) and \(-8\):
\[ (n + 9)(n - 8) = 0 \]
Setting each factor equal to zero gives us:
\[ n + 9 = 0 \quad \Rightarrow \quad n = -9 \] \[ n - 8 = 0 \quad \Rightarrow \quad n = 8 \]
Thus, the two pairs of consecutive integers are:
- \( n = -9 \), \( n + 1 = -8 \)
- \( n = 8 \), \( n + 1 = 9 \)
Therefore, the pairs of consecutive integers that satisfy the condition are:
\((-9, -8)\) and \((8, 9)\).
6 answers