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secondary 4 | E Maths
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Help me with part ii. Thanks in advance.
Date Posted: 4 years ago
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I have written the final expression in the square root as the product of its prime factors.
For the number to be successfully square rooted, the powers of each said prime number must be a multiple of 2. Minimally, we take out one 2, the 5 and the 7. We can take out more 2s if needed, but this will not give the smallest value for p.
So p = 2 x 5 x 7 = 70.
Similarly, for a number to be successfully cube rooted, the powers of each said prime number must be a multiple of 3.
For the number to be successfully square rooted, the powers of each said prime number must be a multiple of 2. Minimally, we take out one 2, the 5 and the 7. We can take out more 2s if needed, but this will not give the smallest value for p.
So p = 2 x 5 x 7 = 70.
Similarly, for a number to be successfully cube rooted, the powers of each said prime number must be a multiple of 3.
Date Posted:
4 years ago
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