Let's denote the two numbers as \( x \) and \( y \). We have the following equations based on your conditions:
- \( x \cdot y = 14 \) (they multiply to give 14)
- \( x + y = 49 \) (they add to give 49)
We can start by expressing \( y \) in terms of \( x \) using the second equation:
\[ y = 49 - x \]
Now we can substitute this expression for \( y \) into the first equation:
\[ x(49 - x) = 14 \]
Expanding this gives:
\[ 49x - x^2 = 14 \]
Rearranging this equation yields:
\[ x^2 - 49x + 14 = 0 \]
Now we can use the quadratic formula to solve for \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation, \( a = 1 \), \( b = -49 \), and \( c = 14 \):
\[ x = \frac{-(-49) \pm \sqrt{(-49)^2 - 4 \cdot 1 \cdot 14}}{2 \cdot 1} \] \[ x = \frac{49 \pm \sqrt{2401 - 56}}{2} \] \[ x = \frac{49 \pm \sqrt{2345}}{2} \]
Now we calculate \( \sqrt{2345} \). The approximate value is around 48.4.
Thus, we find:
\[ x \approx \frac{49 \pm 48.4}{2} \]
Calculating the two possible values for \( x \):
- \( x \approx \frac{49 + 48.4}{2} \approx \frac{97.4}{2} \approx 48.7 \)
- \( x \approx \frac{49 - 48.4}{2} \approx \frac{0.6}{2} \approx 0.3 \)
Now we can find the corresponding values of \( y \) using \( y = 49 - x \):
- If \( x \approx 48.7 \), then \( y \approx 49 - 48.7 = 0.3 \).
- If \( x \approx 0.3 \), then \( y \approx 49 - 0.3 = 48.7 \).
Thus, the two numbers are approximately \( 48.7 \) and \( 0.3 \).
However, since these values do not yield integer solutions and don't seem suitable for the conditions based on standard arithmetic, it's important to clarify that there are no two real numbers that satisfy both conditions exactly.
1 answer