Question

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 1

SOLUTION :

Let the three consecutive integers be x, (x + 1) & (x + 2).

A.T.Q

x² + (x + 1)(x + 2) = 154

x² + x² + 2x + x + 2 = 154

2x² + 3x + 2 = 154

2x² + 3x + 2 - 154 = 0

2x² + 3x - 152 = 0

2x² + 19x - 16x - 152 = 0

[By middle term splitting method]

x(2x + 19) - 8(2x + 19) = 0

(2x + 19)(x - 8) = 0

(2x + 19) = 0  or (x - 8) = 0

2x = -19  or x = 8  

x = -19/2 or x = 8  

Since, x can’t be negative , so x ≠ -19/2.

Therefore, x = 8  

First integer (x) = 8  

Second integer (x + 1) = 8 + 1 =9

Third integer (x + 2) = 8 + 2 = 10  

Hence, the three consecutive integers are 8 ,9 & 10 .

HOPE THIS  ANSWER WILL HELP YOU..

Answer 2

Answer:

Ans : 8, 9, 10

Hint :

Let the integers be x, x + 1 and x + 2.

Then, x

2

+ (x + 1)(x + 2) = 154.

⇒(2x+19)(x−8)