Question

Find two consecutive even numbers whose product is 224

Answer 1

Let's denote the two consecutive even numbers as \( x \) and \( x + 2 \). We know that their product is given by the equation:

\[
x(x + 2) = 224
\]

Expanding this equation, we get:

\[
x^2 + 2x = 224
\]

Next, we can rearrange the equation to set it to zero:

\[
x^2 + 2x - 224 = 0
\]

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

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

In our equation, \( a = 1 \), \( b = 2 \), and \( c = -224 \). Substituting these values into the quadratic formula gives us:

\[
x = \frac{{-2 \pm \sqrt{{2^2 - 4 \cdot 1 \cdot (-224)}}}}{2 \cdot 1}
\]

Calculating the discriminant:

\[
b^2 - 4ac = 4 + 896 = 900
\]

Now substituting back into the formula:

\[
x = \frac{{-2 \pm \sqrt{900}}}{2}
\]

Since \( \sqrt{900} = 30 \), we can continue:

\[
x = \frac{{-2 \pm 30}}{2}
\]

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

1. \( x = \frac{{28}}{2} = 14 \)
2. \( x = \frac{{-32}}{2} = -16 \)

However, since we are looking for positive even numbers, we take \( x = 14 \).

Thus, the two consecutive even numbers are:

\[
14 \quad \text{and} \quad 16
\]

To confirm, we calculate their product:

\[
14 \times 16 = 224
\]

Therefore, the two consecutive even numbers whose product is 224 are \( \boxed{14} \) and \( \boxed{16} \).