This question has been taken from Book:- ML aggarwal, Avichal publication, class10th, quadratic equation in one variable, chapter 5, exercise 5.5

This is an important ques from the chapter.

There are three consecutive positive integers such that the sum of the square of the first and the product of other two is 154. What are the integers?

Question no.7 , ML Aggarwal, chapter 5, exercise 5.5, quadratic equation in one variable, ICSE board,

Let us consider the first integer be ‘x’

Second integer be ‘x + 1’

Third integer be ‘x + 2’

So, according to the question,

x

^{2}+ (x + 1) (x + 2) = 154let us simplify,

x

^{2}+ x^{2}+ 3x + 2 – 154 = 02x

^{2}+ 3x – 152 = 0Let us factorize,

2x

^{2}+ 19x – 16x – 152 = 0x(2x + 19) – 8 (2x + 19) = 0

(2x + 19) (x – 8) = 0

So,

(2x + 19) = 0 or (x – 8) = 0

2x = -19 or x = 8

x = -19/2 or x = 8

∴ The value of x = 8 [since -19/2 is a negative value]

So,

First integer = x = 8

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

Third integer = x + 2 = 8 + 2 = 10

∴ The numbers are 8, 9, 10.