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Question
Find two consecutive multiples of 3 whose product is 648.
Solution
Le the required consecutive multiples of 3 be 3x and`3(x+1)`
According to the given condition,
`3x xx3(x+1)=648`
⇒`9(x^2+x)=648`
⇒` x^2+x=72`
⇒`x^2+x-72=0`
⇒`x^2+x-72=0`
⇒`x^2+9x-8x-72=0`
⇒`x(x+9)-8(x+9)=0`
⇒`(x+9)(x-8)=0`
⇒`x+9=0 or x-8=0`
⇒` x=-9 or x=8`
∴` x=8` (Neglecting the negative value)
When `x=8`
`3x=3xx8=24`
`3(x+1)=3xx(8+1) =3xx9=27`
Hence, the required multiples are 24 and 27.
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