A natural number whose squares diminished by 84 is equal to thrice of 8 more than that number?

Answer 1

A natural number is a counting number - it is positive, 1, 2, 3, 4, and so on.

Let's call our natural number, n.

1. The square of n is the same as saying n²
2. Diminished means made smaller, so we are subtracting 84 from n².
3. 8 more than our number can be represented as n + 8
4. Thrice more than this is 3(n+8)
5. So our equation is n² - 84 = 3(n+8)
6. Simplifying gives us n² - 84 = 3n + 24
7. Let's move all of our terms to the left hand side of our equation,

n² - 3n - 84 - 24 = 0

n² - 3n - 108 = 0

(n - 12)(n + 9) = 0

n - 12 = 0 or n + 9 = 0

n = 12 or n = -9

1. However, -9 is not a natural number.
2. Therefore, our answer is n = 12.

Answer 2

If there is a difficulty in this problem, it is to convert the verbiage of the problem into an equation:

"The square of a number diminished by 84" is just n^2 - 84

"Thrice eight added to the number" is 3 (n + 8)

Now we equate them and move everything to the left side to get a quadratic equation:

n^2 - 84 = 3 (n +8) = 3n + 24, whence n^2 - 84 - 3n - 24=0

Simplifying gives n^2 - 3n -108.

Solving with quadratic equation gives [3+/- sqrt ( 9 +432)]/2=(3 +/- 21)/2. The only one of these two numbers that is a natural ("counting" ) number is (3+21)/2 = 12, so the answer is 12.

Check: For n=12, n^2-84 =144- 84 =60

Also for n=12, 3(n+8)=3*20 =60

So, it checks out.