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+2 x-288=0$
$ \Rightarrow x^2+18 x-16 x-288=0$
$ \Rightarrow x(x+18)-16(x+18)=0$
$ \Rightarrow(x+18)(x-16)=0$
$ \Rightarrow x+18=0 \text { or } x-16=0$
$ \Rightarrow x=-18 \text { or } x=16$
$\text { Since } x \text { is a positive integer, } x \neq-18$
$ \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+2 x-288=0$
$ \Rightarrow x^2+18 x-16 x-288=0$
$ \Rightarrow x(x+18)-16(x+18)=0$
$ \Rightarrow(x+18)(x-16)=0$
$ \Rightarrow x+18=0 \text { or } x-16=0$
$ \Rightarrow x=-18 \text { or } x=16$
$\text { Since } x \text { is a positive integer, } x \neq-18$
$ \Rightarrow x=16$
$ \Rightarrow x+2=16+2=18$
Hence, the required consecutive positive even integers are $16$ and $18.$
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