Question

Find two consecutive integers such that the square of greater is 9 more than square of the smaller.​

Answer 1

Answer:

Q.Find two consecutive integers such that the square of greater is 9 more than square of the smaller.

= Solution;

Let, the consecutive integers be x and x+1

Since, x+1 > x

According to the question,

(x+1)² = x² + 9

or, x²+2x+1 = x² + 9

or, 2x + 1 = 9  [ Eliminating x² from both sides ]

or, 2x = 9-1

or, 2x = 8

or, x = 8÷2

∴ x = 4

∴ x+1 = 4+1=5

Hence, 4 and 5 are the required consecutive integers.

Thank you Rishab for asking