Question

x(x+2)=8
Find the pair(s) of consecutive even integers. If there is more than one pair, use the "or" button.

Answer 1

To solve the equation \( x(x + 2) = 8 \), we can rearrange it as follows:

1. Expand the left side:
\[
x^2 + 2x = 8
\]

2. Move 8 to the left side to set the equation to zero:
\[
x^2 + 2x - 8 = 0
\]

Next, we can use the quadratic formula to find the roots of the equation \( ax^2 + bx + c = 0 \):
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = 2 \), and \( c = -8 \).

3. Calculate the discriminant:
\[
b^2 - 4ac = 2^2 - 4(1)(-8) = 4 + 32 = 36
\]

4. Apply the quadratic formula:
\[
x = \frac{-2 \pm \sqrt{36}}{2(1)} = \frac{-2 \pm 6}{2}
\]
Which gives us two solutions:
- \( x = \frac{4}{2} = 2 \)
- \( x = \frac{-8}{2} = -4 \)

Now we have the values of \( x = 2 \) and \( x = -4 \).

The consecutive even integers corresponding to these values of \( x \) are:

- For \( x = 2 \): The consecutive even integers are \( 2 \) and \( 4 \).
- For \( x = -4 \): The consecutive even integers are \( -4 \) and \( -2 \).

Thus, the pairs of consecutive even integers we have are:

\[
(2, 4) \quad \text{or} \quad (-4, -2)
\]