Question

three consecutive natural numbers are such that the square of the middle number exceeds the difference of the square of the other two by 60 .find the numbers

Answer 1

Let the three consecutive natural numbers be x, x+1, x+2

Given that the square of the middle number exceeds the difference of the square of the other two numbers by 60
Therefore
Difference between the square of the first number and the third number
= (x + 2)² - x²
= x² + 4 + 4x - x²
= 4x + 4
According to the sum
(x + 1)² = 4x + 4 + 60
x² + 1 + 2x = 4x + 64
x² + 2x - 4x + 1 - 64 = 0
x² - 2x - 63 = 0
x² - 9x + 7x - 63 = 0
x (x - 9) + 7(x - 9) = 0
(x - 9) (x + 7) = 0
x = 9 or x = - 7
Since x is a natural we have to reject x = -7

Therefore x = 9
x + 1 = 10
x + 2 = 11

The three natural numbers are 9, 10, 11.

Hope my answer helps you.

Answer 2

Step-by-step explanation:

Let the three consecutive natural numbers be x,x+1, x+2.

Given that Square of the middle number exceeds the difference of the squares of the other two by 60.

(x + 1)^2 = (x + 2)^2 - (x)^2 + 60

x^2 + 1 + 2x = x^2 + 4 + 4x - x^2 + 60

x^2 + 1 + 2x = 4x+ 64

x^2 = 4x + 64 - 2x - 1

x^2 = 2x + 63

x^2 - 2x - 63 = 0

x^2 - 9x + 7x - 63 = 0

x(x - 9) + 7(x - 9) = 0

(x - 9)(x + 7) = 0

x = 9,-7.

x value cannot be -ve, so, x = 9.

Then,

x + 1 = 10

x + 2 = 11.

Therefore the three natural numbers are 9,10,11.

Verification:

(x + 1)^2 - (x + 2)^2 + x^2 = 60

10^2 - 11^2 + 9^2 = 60

100 - 121 + 81 = 60

-21 + 81 = 60

60 = 60.

Hope this helps!