Question 204320
<font size = 7 "color = "red"><b>Edwin's solution and explanation:
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Three times the square of a certain postive number exceeds
six times the number by nine.
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Let N be "a certain positive number" as well as "the number"  
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Three times the square of N exceeds six times N by nine.
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Let "the square of N" be N<sup>2</sup>.
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Three times N<sup>2</sup> exceeds six times N by nine.
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Replace "Three times N<sup>2</sup> by 3N<sup>2</sup>.
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3N<sup>2</sup> exceeds six times N by nine.
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Replace "six times N" by 6N
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3N<sup>2</sup> exceeds 6N by nine.
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That means if you subtract them (larger minus smaller)
you'll get 9.

The larger is the one that does the exceeding, 3N<sup>2</sup>,
so the smaller one is 6N, so when we subtract them by putting
a minus sign between them, we must get 9.
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       3N<sup>2</sup> - 6N = 9

   3N<sup>2</sup> - 6N - 9 = 0

 3(N<sup>2</sup> - 2N - 3) = 0

3(N - 3)(N + 1) = 0
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Divide both sides by 3
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 (N - 3)(N + 1) = 0

N - 3 = 0;    N + 1 = 0

    N = 3;        N = -1
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Since the requirement "a certain POSITIVE numbers"
was given we discared the -1 and the only solution
is N = 3

Checking:

Three times the square of 3 exceeds six times 3 by nine.

Three times 9 exceeds six times 3 by nine.

27 exceeds 6 times 3 by 9

27 exceeds 18 by 9

That is true.

Edwin</pre>