Question

Use a quadratic equation to find two real numbers that satisfy each situation or show that no such numbers exist.
Their sum is 15 and their product is 36

Answer 1

To solve for two real numbers whose sum is 15 and whose product is 36, we can set up a system of equations based on the given information.

Let the two numbers be \( x \) and \( y \).

We know:
1. \( x + y = 15 \) (Equation 1)
2. \( x \cdot y = 36 \) (Equation 2)

From Equation 1, we can express \( y \) in terms of \( x \):
\[ y = 15 - x \]

Now, we can substitute this expression for \( y \) into Equation 2:
\[
x(15 - x) = 36
\]

Expanding this, we get:
\[
15x - x^2 = 36
\]
Rearranging it into standard quadratic form:
\[
-x^2 + 15x - 36 = 0
\]
Multiplying through by -1 to standardize:
\[
x^2 - 15x + 36 = 0
\]

Next, we will apply the quadratic formula, which is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In our equation, \( a = 1 \), \( b = -15 \), and \( c = 36 \).

First, we calculate the discriminant:
\[
b^2 - 4ac = (-15)^2 - 4(1)(36) = 225 - 144 = 81
\]

Since the discriminant is positive, we will have two distinct real solutions.

Calculating the roots:
\[
x = \frac{15 \pm \sqrt{81}}{2(1)} = \frac{15 \pm 9}{2}
\]

This gives us two possible values for \( x \):
\[
x_1 = \frac{15 + 9}{2} = \frac{24}{2} = 12
\]
\[
x_2 = \frac{15 - 9}{2} = \frac{6}{2} = 3
\]

Thus, the two numbers are:
- \( x = 12 \)
- \( y = 3 \)

Check:
- The sum \( 12 + 3 = 15 \) (correct)
- The product \( 12 \cdot 3 = 36 \) (correct)

Therefore, the two numbers that satisfy the conditions are **12 and 3**.