There are three consecutive integers such that the square of the first

1.	There are three consecutive integers such that the square of the first increased by the product of the other two gives 154. What are the integers?
Answer» Let’s consider the three consecutive numbers to be x, x + 1, x + 2 respectively. And x being the first integer of the sequence. From the question, it’s understood that x²+ (x + 1)(x + 2) = 154 ⇒ x²+ x²+ 3x + 2 = 154 ⇒ 2x²+ 3x – 152 = 0 Solving for x by factorization method, we have ⇒ 2x²+ 19x – 16x – 152 = 0 ⇒ x(2x + 19) – 8(2x – 19) = 0 ⇒ (2x – 19)(x – 8) = 0 Now, either 2x – 19 = 0 ⇒ x = 19/2 (which is not an integer) Or, x – 8 = 0 ⇒ x = 8 Hence, considering x = 8 the three consecutive integers are 8, 9 and 10.

There are three consecutive integers such that the square of the first increased by the product of the other two gives 154. What are the integers?

Answer»

Let’s consider the three consecutive numbers to be x, x + 1, x + 2 respectively. And x being the first integer of the sequence.

From the question, it’s understood that

x²+ (x + 1)(x + 2) = 154

⇒ x²+ x²+ 3x + 2 = 154

⇒ 2x²+ 3x – 152 = 0

Solving for x by factorization method, we have

⇒ 2x²+ 19x – 16x – 152 = 0

⇒ x(2x + 19) – 8(2x – 19) = 0

⇒ (2x – 19)(x – 8) = 0

Now, either 2x – 19 = 0 ⇒ x = 19/2 (which is not an integer)

Or, x – 8 = 0 ⇒ x = 8

Hence, considering x = 8 the three consecutive integers are 8, 9 and 10.