Question

5. There are three consecutive integers such thi
square of the first increased by the product
other two gives 154. Find the integers.
​

Answer 1

Given :

• Three consecutive positive integer are such that the sum of the square of the first and the product of their two integers is 154.

To find :

• The integers =?

Step-by-step explanation :

Let, the first consecutive positive integer be x.

Then, the second consecutive positive integer be x + 1.

And, the third consecutive positive integer be x + 2.

It is Given that,

The sum of the square of the first and the product of their two integers is 154.

According to the question :

➟ x² + (x + 1)(x + 2) = 154

➟ x² + x² + 3x + 2 = 154

➟ 2x² + 3x - 152 = 0

➟ 2x² - 16x + 19x - 152 = 0

➟ 2x (x - 8) + 19(x - 8) = 0

➟ (2x + 19)(x - 8) = 0

Now,

Value of x,

2x + 19 = 0

2x = - 19

x = - 19/2 [Ignore Negative]

Or,

x - 8 = 0

x = 8.

Now,

x = - 19/2 Or x = 8.

We can't take the negative values. So, x = 8.

Therefore,

The first consecutive positive integer, x = 8

Then, the second consecutive positive integer, x + 1 = 8 + 1 = 9

And, the third consecutive positive integer, x + 2 = 8 + 2 = 10

Answer 2

Let the three consecutive integers be "x" , "x+1" , "x+2"

A/c , " square of the first increased by the product of other two gives 154 "

⇒ (x)² + (x+1)(x+2) = 154

⇒ x² + x² + 2x + x + 2 = 154

⇒ 2x² + 3x - 152 = 0

⇒ 2x² - 16x + 19x - 152 = 0

⇒ 2x ( x-8 ) + 19 ( x-8 ) = 0

⇒ ( x-8 ) ( 2x+19 ) = 0

⇒ x = 8

Ignore negative decimal value as it is not an integer .

i.e., x ≠ -19/2

So three consecutive integers are 8 , (8+1) , (8+2) .

So 8 , 9 , 10 are three consecutive integers.