Question

7./ Alice and Felix think of two numbers. The numbers are 10 apart and their product is 56.
Solve the equation x(x- l0) = 56 to work out the numbers Alice and Felix thought of
(there are two different possible pairs of numbers).
has

Answer 1

Let's denote the two numbers that Alice and Felix thought of as \( x \) and \( y \). According to the problem:

1. The two numbers are 10 apart, which gives us the equation:
\[
y = x - 10
\]

2. 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:
1. \( x = \frac{10 + 18}{2} = \frac{28}{2} = 14 \)
2. \( x = \frac{10 - 18}{2} = \frac{-8}{2} = -4 \)

Now we can find the corresponding \( y \) values using \( y = x - 10 \):
1. For \( x = 14 \):
\[
y = 14 - 10 = 4
\]
Thus, one pair of numbers is \( (14, 4) \).

2. 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:
\[
(14, 4) \quad \text{and} \quad (-4, -14)
\]