Question

Play the game! Find two numbers that multiply to make the 1st number and add to make the 2nd number. Multiply 108; Add -31

Answer 1

Let's denote the two numbers as \( x \) and \( y \). According to the problem, we have the following two equations:

1. \( x \cdot y = 108 \) (they multiply to make 108)
2. \( x + y = -31 \) (they add to make -31)

We can solve this system of equations step-by-step.

From the second equation, we can express \( y \) in terms of \( x \):

\[
y = -31 - x
\]

Now we can substitute this expression for \( y \) into the first equation:

\[
x \cdot (-31 - x) = 108
\]

Expanding this, we have:

\[
-31x - x^2 = 108
\]

Rearranging this gives us a standard quadratic equation:

\[
x^2 + 31x + 108 = 0
\]

Next, we can use the quadratic formula to solve for \( x \):

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In our case, \( a = 1 \), \( b = 31 \), and \( c = 108 \). Plugging these values into the formula:

\[
x = \frac{-31 \pm \sqrt{31^2 - 4 \cdot 1 \cdot 108}}{2 \cdot 1}
\]

Calculating the discriminant:

\[
31^2 = 961
\]
\[
4 \cdot 1 \cdot 108 = 432
\]
\[
31^2 - 4 \cdot 1 \cdot 108 = 961 - 432 = 529
\]

Now we find the square root of the discriminant:

\[
\sqrt{529} = 23
\]

Now we substitute back into the formula:

\[
x = \frac{-31 \pm 23}{2}
\]

This gives us two possible solutions for \( x \):

1. \( x = \frac{-31 + 23}{2} = \frac{-8}{2} = -4 \)
2. \( x = \frac{-31 - 23}{2} = \frac{-54}{2} = -27 \)

Now, we can find the corresponding \( y \) values:

1. If \( x = -4 \):
\[
y = -31 - (-4) = -31 + 4 = -27
\]

2. If \( x = -27 \):
\[
y = -31 - (-27) = -31 + 27 = -4
\]

Thus, the two numbers are \( -4 \) and \( -27 \).

To confirm:
- Their product: \( -4 \cdot -27 = 108 \) (correct)
- Their sum: \( -4 + (-27) = -31 \) (correct)

So the two numbers are **-4 and -27**.