Let the two numbers be \( x \) and \( y \), where \( x \leq y \). We are given that: \[ x + y = 25 \quad and \quad x \times y = 144 \] Use the quadratic equation: \[ t^2 - (x + y)t + x \times y = 0 \] \[ t^2 - 25t + 144 = 0 \] Solve this quadratic equation: \[ t = \frac{25 \pm \sqrt{25^2 - 4 \times 1 \times 144}}{2 \times 1} = \frac{25 \pm \sqrt{625 - 576}}{2} = \frac{25 \pm \sqrt{49}}{2} \] \[ t = \frac{25 \pm 7}{2} \] Thus, \( t = 9 \) or \( t = 16 \). Therefore, the smaller number is 9. To find two numbers with a given sum and product, solve the corresponding quadratic equation.