Let's denote the two numbers as \( x \) and \( y \). According to the problem, we have the following equations:
- \( x + y = 22 \) (the sum of the numbers)
- \( xy = 120 \) (the product of the numbers)
We can express \( y \) in terms of \( x \) using the first equation:
\[ y = 22 - x \]
Now, we can substitute this expression for \( y \) into the second equation:
\[ x(22 - x) = 120 \]
Expanding this, we get:
\[ 22x - x^2 = 120 \]
Rearranging gives us a standard form quadratic equation:
\[ x^2 - 22x + 120 = 0 \]
To solve for \( x \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation, \( a = 1 \), \( b = -22 \), and \( c = 120 \). Substituting these values in:
\[ x = \frac{22 \pm \sqrt{(-22)^2 - 4 \cdot 1 \cdot 120}}{2 \cdot 1} \] \[ x = \frac{22 \pm \sqrt{484 - 480}}{2} \] \[ x = \frac{22 \pm \sqrt{4}}{2} \] \[ x = \frac{22 \pm 2}{2} \]
This gives us two possible solutions for \( x \):
- \( x = \frac{24}{2} = 12 \)
- \( x = \frac{20}{2} = 10 \)
Thus, the two numbers \( x \) and \( y \) are \( 10 \) and \( 12 \). Since we want to input the lesser number first, the answer is:
10 and 12.
1 answer