the sum of squares of 2 consecutive numbers is 145, what are the numbers?

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

The sum of the squares of these numbers is given by:
\[
x^2 + (x+1)^2 = 145
\]

First, expand the equation:
\[
x^2 + (x^2 + 2x + 1) = 145
\]

Combine like terms:
\[
2x^2 + 2x + 1 = 145
\]

Subtract 145 from both sides to set the equation to zero:
\[
2x^2 + 2x + 1 - 145 = 0
\]
\[
2x^2 + 2x - 144 = 0
\]

Divide the entire equation by 2 to simplify:
\[
x^2 + x - 72 = 0
\]

Next, solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 \), and \( c = -72 \).

Calculate the discriminant:
\[
b^2 - 4ac = 1^2 - 4(1)(-72) = 1 + 288 = 289
\]

Find the square root of the discriminant:
\[
\sqrt{289} = 17
\]

Apply the quadratic formula:
\[
x = \frac{-1 \pm 17}{2}
\]

This gives us two solutions:
\[
x = \frac{-1 + 17}{2} = \frac{16}{2} = 8
\]
\[
x = \frac{-1 - 17}{2} = \frac{-18}{2} = -9
\]

Therefore, the two pairs of consecutive numbers are:
1. \( x = 8 \) and \( x+1 = 9 \)
2. \( x = -9 \) and \( x+1 = -8 \)

So, the numbers are either \( (8, 9) \) or \( (-9,

Explore how to solve the problem where the sum of squares of two consecutive numbers equals 145. Step-by-step solution using algebraic methods, including the quadratic equation formula. Suitable for Grades 7-9 students.

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