SOLUTION: 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

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 =  = 



Middle number = n+1

Square of the middle number = 



Smallest number = n

3 times the smallest number = 



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



 = 



This value "exceeds '3 times the smallest' by 7" means this is equal to 

"3 times the smallest + 7" i.e. it is 



Equating the above, 



i.e. 

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!



:)



 
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation  (in our case ) has the following solutons:

  

  

  

  For these solutions to exist, the discriminant  should not be a negative number.

  

  First, we need to compute the discriminant : .

  

  Discriminant d=100 is greater than zero. That means that there are two solutions: .

  

    

    

    

    Quadratic expression  can be factored:

  

  Again, the answer is: 8, -2.
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