Question

Determine the sum of two consecutive multiples of $3$ whose product is $270$.

Answer 1

$\Rightarrow$ Let the two multiples be $x$ and $x+3$

According to the question,

$\Rightarrow$ $x(x+3)=270$

$\Rightarrow$ $x^2+3x-270=0$

$\Rightarrow$ $x^2+18x-15x-270=0$

$\Rightarrow$ $x(x+18)-15(x+18)=0$

$\Rightarrow$ $(x+18)(x-15)=0$

$\Rightarrow$ $x+18=0$ and $x-15=0$

$\Rightarrow$ $x=-18$ and $x=15$

$\Rightarrow$ $x$ cannot be negative, so $x=15$

$\Rightarrow$ Other number $x+3=15+3=18$

$\Rightarrow$ Sum of the numbers $=15+18=33$