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 find the numbers

Answer 1

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

Given that x² exceeds the difference of the squares of x - 1 and x + 1 by 60.

So the difference between x² and (x + 1)² - (x - 1)² is 60.

$x^2-((x+1)^2-(x-1)^2)=60 \\ \\ (x+1)^2-(x-1)^2=x^2-60 \\ \\ 4x=x^2-60 \\ \\ x^2-4x-60=0 \\ \\ x^2-10x+6x-60=0 \\ \\ x(x-10)+6(x-10)=0 \\ \\ (x+6)(x-10) \\ \\ \\ \therefore\ x=-6 \ \ \ ; \ \ \ x = 10$

As the three are natural numbers, x ≠ - 6.

∴ x = 10

x - 1 = 9

x + 1 = 11

∴ 9, 10, 11 are the numbers.

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Answer 2

CHECK THE ATTACHMENT FOR SOLUTION.

HOPE IT HELPS.