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, $x \neq -18$
$\Rightarrow x = 16$
$\Rightarrow x + 2 = 16 + 2 = 18$
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, $x \neq -18$
$\Rightarrow x = 16$
$\Rightarrow x + 2 = 16 + 2 = 18$
Hence, the required consecutive positive even integers are $16$ and $18$.
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