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