Three consecutive positive integers are such that sum of square of the first and the product of the other two is 67, find the integers.
Let the three consecutive positive integers be n, n+1, n+2
Given that
that sum of square of the first and the product of the other two is 67
n² + (n+1)(n+2) = 67
n² + (n² + 2n + n + 2) = 67
n² + n² + 3n + 2 = 67
2n² + 3n + 2 = 67
2n² + 3n + 2 - 67 = 0
2n² + 3n − 65 = 0
Solve by quadratic formula or factorisation
Factorisation method
Find two numbers that gives a product of 2 × (−65) = −130 and add to 3.
by trial and error the numbers13 and −10 satisfy the condition
use this to split the middle term of
2n² + 3n − 65 = 0
.2n² + 13n − 10n − 65 = 0
n(2n + 13) − 5(2n + 13) = 0
(n − 5)(2n + 13) = 0
solving
n − 5 = 0 ⇒ n = 5
2n + 13 = 0 ⇒ n = −13/2 (rejecct)
required positive integers are
n=5
n+1 =6
n+2=7
for a video of the solution
cbse 10th maths pervious years old question papers 2025 2026 standard mathematics
quadratic equation word problem formation and solution by splitting the middle term
