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InterviewSolution
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Find two consecutive multiples of 3 whose product is 648 |
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Answer» Let the\xa0consecutive multiples of 3 be 3x and 3(x + 1).Then, we have\xa0{tex}3x\\times3(x + 1) = 648{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}9x^2\xa0+ 9x - 648 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x^2\xa0+ x - 72 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x^2\xa0+ 9x - 8x - 72 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x(x + 9) - 8(x + 9) = 0{/tex}{tex}\\Rightarrow{/tex}\xa0x + 9\xa0= 0 or x - 8 = o{tex}\\Rightarrow{/tex}\xa0x = -9 or x = 8Since x is a positive number, x\xa0{tex}\\neq{/tex}\xa0-9{tex}\\Rightarrow{/tex}\xa0x = 8{tex}\\Rightarrow{/tex}\xa03x = 3(8) = 24 and 3(x + 1) = 3(9) = 27Hence, the required consecutive multiples of 3 are 24 and 27. Language is strange of Sia so I am not understanding please use simple language |
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