Question 766392
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Question:
find the smallest of 3 consecutive positive integers such that when 5 times the largest is subtracted from the square of the middle one the result exceeds three times the smallest by 7.
Ans:

Let the numbers be n and (n+1) and (n+2)

(You can also assume n-1, n and n+1 - the approach is still the same)

Now let's look at each part of the question and translate it into algebraic form.

Largest number = n+2
5 times the largest number = {{{5*(n+2)}}} = {{{5*n + 10}}}

Middle number = n+1
Square of the middle number = {{{(n+1)^2 = n^2 + 2*n + 1}}}

Smallest number = n
3 times the smallest number = {{{3*n}}}

"5 times largest number" subtracted from "square of middle number" will give you

{{{n^2 + 2*n + 1 - (5*n + 10)}}} = {{{n^2 - 3*n -9}}}

This value "exceeds '3 times the smallest' by 7" means this is equal to 
"3 times the smallest + 7" i.e. it is {{{3*n + 7}}}

Equating the above, {{{n^2 - 3*n - 9 = 3*n + 7}}}

i.e. {{{n^2 -6*n - 16 = 0}}}
This is a standard quadratic equation, which can be solved by factorization or
by the quadratic formula, as shown below.

The solutions are n = 8, or n = -2

Since it is given that the numbers are positive, the set of 3 numbers are
8,9 and 10.

Check: Square of middle number - 5 times largest number = 81 - 50 = 31
3 times smallest number = 3 * 8 = 24.
31 exceeds 24 by 7 - correct!

:)
*[invoke quadratic "x", 1, -6, -16 ]