Question

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

Answer 1

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Answer 2

Step-by-step explanation:

Given :

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

To find :

The three consecutive natural numbers.

Solution :

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

According to the question :

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

⇒ (x + 1)² = (x + 2)² - (x)² + 60

⇒ x² + 2x + 1 = x² + 4x + 4 - x² + 60

⇒ x² - 2x - 63 = 0

Solving the quadratic equation :

⇒ x² - 9x + 7x - 63 = 0

⇒ x(x - 9) + 7(x - 9) = 0

⇒ (x + 7) (x - 9) = 0

⇒ x = -7, 9  [x ≠ -'ve]

∴ x = 9

The numbers are :

1st number  (x) = 9

2nd number (x + 1)= 9 + 1 = 10  

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

∴ The three consecutive 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
• 100 = 40 + 60
• 100 = 100

∴ L.H.S = R.H.S

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