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.

 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

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