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