Question
Find two consecutive positive even integers whose product is 288.
✓
Answer
Let the required consecutive positive even integers be x and (x + 2).
Then, we have
$x \times (x + 2) = 288$
$\Rightarrow x^2 + 2x - 288 = 0$
$\Rightarrow x^2 + 18x - 16x - 288 = 0$
$\Rightarrow x(x + 18) - 16(x + 18) = 0$
$\Rightarrow (x + 18)(x - 16) = 0$
$\Rightarrow x + 18 = 0 or x - 16 = 0$
$\Rightarrow x = -18 or x = 16$
Since x is a positive integer, x ≠ -18
⇒ x = 16
⇒ x + 2 = 16 + 2 = 18
Hence, the required consecutive positive even integers are 16 and 18.
Then, we have
$x \times (x + 2) = 288$
$\Rightarrow x^2 + 2x - 288 = 0$
$\Rightarrow x^2 + 18x - 16x - 288 = 0$
$\Rightarrow x(x + 18) - 16(x + 18) = 0$
$\Rightarrow (x + 18)(x - 16) = 0$
$\Rightarrow x + 18 = 0 or x - 16 = 0$
$\Rightarrow x = -18 or x = 16$
Since x is a positive integer, x ≠ -18
⇒ x = 16
⇒ x + 2 = 16 + 2 = 18
Hence, the required consecutive positive even integers are 16 and 18.
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