Let's denote the two numbers that Alice and Felix thought of as \( x \) and \( y \). According to the problem:
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The two numbers are 10 apart, which gives us the equation: \[ y = x - 10 \]
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The product of the two numbers is 56, leading to the equation: \[ x \cdot y = 56 \]
Substituting the first equation into the second, we get: \[ x(x - 10) = 56 \]
Expanding the equation: \[ x^2 - 10x = 56 \]
Now, we'll rearrange it into standard quadratic form: \[ x^2 - 10x - 56 = 0 \]
Next, we can solve for \( x \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -10 \), and \( c = -56 \).
Calculating the discriminant: \[ b^2 - 4ac = (-10)^2 - 4(1)(-56) = 100 + 224 = 324 \]
Now calculate \( x \): \[ x = \frac{-(-10) \pm \sqrt{324}}{2 \cdot 1} = \frac{10 \pm 18}{2} \]
This gives us the two possible values:
- \( x = \frac{10 + 18}{2} = \frac{28}{2} = 14 \)
- \( x = \frac{10 - 18}{2} = \frac{-8}{2} = -4 \)
Now we can find the corresponding \( y \) values using \( y = x - 10 \):
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For \( x = 14 \): \[ y = 14 - 10 = 4 \] Thus, one pair of numbers is \( (14, 4) \).
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For \( x = -4 \): \[ y = -4 - 10 = -14 \] Thus, the second pair is \( (-4, -14) \).
The two different pairs of numbers that Alice and Felix thought of are: \[ (14, 4) \quad \text{and} \quad (-4, -14) \]