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Find two consecutive multiples of 3 …

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Find two consecutive multiples of 3 whose product is 648

Posted by Nazish Parween 6 years, 5 months ago

CBSE > Class 10 > Mathematics

2 answers

Sia ? 6 years, 5 months ago

Let the consecutive multiples of 3 be 3x and 3(x + 1).
Then, we have
{tex}3x\times3(x + 1) = 648{/tex}
{tex}\Rightarrow{/tex} {tex}9x^2 + 9x - 648 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}x^2 + x - 72 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}x^2 + 9x - 8x - 72 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}x(x + 9) - 8(x + 9) = 0{/tex}
{tex}\Rightarrow{/tex} x + 9 = 0 or x - 8 = o
{tex}\Rightarrow{/tex} x = -9 or x = 8
Since x is a positive number, x {tex}\neq{/tex} -9
{tex}\Rightarrow{/tex} x = 8
{tex}\Rightarrow{/tex} 3x = 3(8) = 24 and 3(x + 1) = 3(9) = 27
Hence, the required consecutive multiples of 3 are 24 and 27.

0Thank You

Iron Man 6 years, 5 months ago

Language is strange of Sia so I am not understanding please use simple language

0Thank You

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