Question

three consecutive natural numbers ae such that the square of the middle number exceeds the difference of the squares of the other two numbers by 60

Answer 1

the three consecutive natural numbers are 9.10 and 11

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!