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

Since the middle number of the three consecutive numbers is x, the other two numbers are x -1 and x +1.
According to the given condition, we have
x² = [(x +1)² -(x -1)²] +60
x² = (x² +2x +1) -(x² -2x +1) +60 = 4x +60
x² -4x -60 = 0 => (x -10)(x +6) = 0
x = 10 or -6
Since x is a natural number, we get x = 10
Hence the three numbers are 9, 10, 11

Answer 2

Answer:

Step-by-step explanation:

Solution :-

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

According to the Question,

⇒ (x + 1)² = (x + 2)² - (x)² + 60

⇒ x² + 2x + 1 = x² + 4x + 4 - x² + 60

⇒ x² - 2x - 63 = 0

⇒ x² - 9x + 7x - 63 = 0

⇒ x(x - 9) + 7(x - 9) = 0

⇒ (x - 9) (x + 7) = 0

⇒ x = 9, - 7 (As x can't be negative)

⇒ x = 9

1st number = x = 9

2nd number = x + 1 = 9 + 1 = 10

3rd number = x + 2 = 9 + 2 = 11

Hence, the three numbers are 9, 10 and 11.