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 96. Assume the middle number to be x and form a quadratic equation satisfying the above statement. Solve the equation and hence find the three numbers.

Answer 1

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.

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