find two positive consecutive integers such that the squares of the first, added to 3 times the second, is 24.

Let the two consecutive positive integers be \( x \) and \( x+1 \). According to the problem:

- The square of the first integer is \( x^2 \).
- Three times the second integer is \( 3(x+1) \).

The sum of these two expressions is equal to 24:
\[
x^2 + 3(x+1) = 24
\]

### Solving the equation:
1. Expand the equation:
   \[
   x^2 + 3(x+1) = x^2 + 3x + 3 = 24
   \]

2. Subtract 24 from both sides:
   \[
   x^2 + 3x + 3 - 24 = 0
   \]
   \[
   x^2 + 3x - 21 = 0
   \]

3. Solve the quadratic equation using the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   Here, \( a = 1 \), \( b = 3 \), and \( c = -21 \). Substituting into the formula:
   \[
   x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-21)}}{2(1)}
   \]
   \[
   x = \frac{-3 \pm \sqrt{9 + 84}}{2}
   \]
   \[
   x = \frac{-3 \pm \sqrt{93}}{2}
   \]

Since \( \sqrt{93} \) is not a perfect square, we cannot get a clean integer solution from this quadratic formula.

### Checking simpler integer values:
We simplify our approach by testing small positive integers directly. Let’s test consecutive integers manually:
1. For \( x = 3 \):
   - First integer squared: \( 3^2 = 9 \)
   - Three times the second integer: \( 3(4) = 12 \)
   - Sum: \( 9 + 12 = 21 \), not 24.
   
2. For \( x = 4 \):
   - First integer squared: \( 4^2 = 16 \)
   - Three times the second integer: \( 3(5) = 15 \)
   - Sum: \( 16 + 15 = 31 \), not 24.

Thus, none of the two initial

Find two positive consecutive integers such that the squares of the first, added to 3 times the second, is 24.

Learn how to find two positive consecutive integers such that the square of the first integer plus three times the second equals 24. This problem explores quadratic equations and integer solutions. Perfect for students in Grades 8-10.

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