Find two number the have a sum of 23 and a product of 120

Let's denote the two numbers as \( x \) and \( y \).

We are given the following conditions:

\[
x + y = 23
\]
\[
x \times y = 120
\]

To find \( x \) and \( y \), we can solve this system of equations. 

First, express \( y \) in terms of \( x \) using the first equation:

\[
y = 23 - x
\]

Substitute this into the second equation:

\[
x \times (23 - x) = 120
\]

This expands to:

\[
23x - x^2 = 120
\]

Rearrange the equation into a standard quadratic form:

\[
x^2 - 23x + 120 = 0
\]

Now, we solve this quadratic equation using the quadratic formula:

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

For our equation \( x^2 - 23x + 120 = 0 \), the coefficients are \( a = 1 \), \( b = -23 \), and \( c = 120 \). Plugging these into the quadratic formula:

\[
x = \frac{-(-23) \pm \sqrt{(-23)^2 - 4(1)(120)}}{2(1)}
\]

\[
x = \frac{23 \pm \sqrt{529 - 480}}{2}
\]

\[
x = \frac{23 \pm \sqrt{49}}{2}
\]

\[
x = \frac{23 \pm 7}{2}
\]

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

\[
x = \frac{23 + 7}{2} = 15 \quad \text{or} \quad x = \frac{23 - 7}{2} = 8
\]

Thus, the two numbers are \( 15 \) and \( 8 \).

Let's verify:
- Their sum: \( 15 + 8 = 23 \)
- Their product: \( 15 \times 8 = 120 \)

Both conditions are satisfied, so the two numbers are \( \boxed{15 \text{ and } 8} \).

Would you like more details on the solution or have any other questions?

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Here are five related questions to expand on this topic:

1. How can you solve a system of linear and quadratic equations?
2. What are the different methods to solve a quadratic equation?
3. Can you always factorize a quadratic equation?
4. What is the significance of the discriminant in the quadratic formula?
5. How do you solve a quadratic equation when the discriminant is negative?

**Tip:** When solving quadratic equations, always check if the equation can be factored easily before using the quadratic formula. This can save time and simplify the process.

Learn how to find two numbers that sum to 23 and have a product of 120 using algebraic methods. This problem involves solving a system of equations and understanding the quadratic formula. Suitable for high school students.

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