Question
Determine two consecutive multiples of $3$ whose product is $270.$
✓
Answer
Let first multiple of $3 = 3x$
Then second multiple of $3 = 3x + 3$
According to the condition,
$3 x(3 x+3)=270$
$\Rightarrow 9 x^2+9 x-270=0 $
$\left.\Rightarrow x^2+x-30=0 \text { (Dividing by } 9\right)$
$\Rightarrow x^2+6 x-5 x-30=0 $
$\Rightarrow x(x+6)-5(x+6)=0 $
$ \Rightarrow(x+6)(x-5)=0$
$\text { Either } x+6=0 \text {, then } x=-6$
Or $x - 5 = 0,$ then $x = 5$
Hence number are $15, 18$ or $-18, -15$
Then second multiple of $3 = 3x + 3$
According to the condition,
$3 x(3 x+3)=270$
$\Rightarrow 9 x^2+9 x-270=0 $
$\left.\Rightarrow x^2+x-30=0 \text { (Dividing by } 9\right)$
$\Rightarrow x^2+6 x-5 x-30=0 $
$\Rightarrow x(x+6)-5(x+6)=0 $
$ \Rightarrow(x+6)(x-5)=0$
$\text { Either } x+6=0 \text {, then } x=-6$
Or $x - 5 = 0,$ then $x = 5$
- When $x = -6,$ then
- If $x = 5,$ then
Hence number are $15, 18$ or $-18, -15$
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