the sum of the square of two consecutive integers is 145. find the integers

Let the two consecutive integers be \( x \) and \( x+1 \).

The sum of the square of these two consecutive integers is given by the equation:

\[
x^2 + (x+1)^2 = 145
\]

Now expand the equation:

\[
x^2 + (x^2 + 2x + 1) = 145
\]

Simplify it:

\[
2x^2 + 2x + 1 = 145
\]

Subtract 145 from both sides to set the equation to 0:

\[
2x^2 + 2x + 1 - 145 = 0
\]

\[
2x^2 + 2x - 144 = 0
\]

Now, divide the whole equation by 2 to simplify it:

\[
x^2 + x - 72 = 0
\]

This is a quadratic equation. To solve it, we can factor it:

\[
(x + 9)(x - 8) = 0
\]

So, \( x = -9 \) or \( x = 8 \).

Thus, the two pairs of consecutive integers are:
- \( x = -9 \), \( x+1 = -8 \)
- \( x = 8 \), \( x+1 = 9 \)

Therefore, the integers are either \( (-9, -8) \) or \( (8, 9) \).

Would you like a detailed explanation of any part of the solution?

Here are five questions that build on this problem:
1. How do you factor a quadratic equation?
2. Can the quadratic formula be used here instead of factoring?
3. What if the sum of squares was different, say 169, how would the solution change?
4. How does completing the square work for this problem?
5. Could there be any other pairs of consecutive numbers?

**Tip:** Always check your solutions by substituting them back into the original equation!

The sum of the square of two consecutive integers is 145. Find the integers.

Learn how to find two consecutive integers whose squares sum to 145 using algebraic methods. This step-by-step guide covers expanding and factoring quadratic equations, making it suitable for students in Grades 8-10.

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