Question

Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46, find the integers.
Or
The difference of squares of two numbers is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers.

Answer 1

Let the numbers are x,x+1,x+2.
As per the question,
x^2 +(x+1)(x+2) = 46.
i.e. 2x^2 +3x-44 = 0.
Now, By applying quadratic formula, you will get
x= -22/4, 4.
But x can't be negative.
So, Numbers will be 4,5 and 6.

Answer 2

Let the three consecutive positive integers be x, x + 1, x + 2
According to the given condition,
x2 + (x + 1)(x + 2) = 46
x2 + x2 + 3x + 2 = 46
2x2 + 3x – 44 = 0
Using the splitting middle term – the middle term of the general equation is divided in two such values that:
Product = a.c
For the given equation a = 2 b = 3 c = – 44
= 2. – 44 = – 88
And either of their sum or difference = b
= 3
Thus the two terms are 11 and – 8
Sum = 11 – 8 = 3
Product = 11. – 8 = – 88
2x2 + 3x – 44 = 0
2x2 + 11x – 8x – 44 = 0
x(2x + 11) – 4(2x + 11) = 0
(2x + 11)(x – 4) = 0
x = 4 or – 11/2
x = 4 (x is a positive integers)
When x = 4
x + 1 = 4 + 1 = 5
x + 2 = 4 + 2 = 6
Hence the required integers are 4, 5, 6