Question

Three consecutive numbers are such that the square of the middle number exceed the difference of the square of the the other two by 60 find number?

Answer 1

Solution:-

Let the Three Consecutive Numbers be "x-1", "x" and "x+1".

A.T.Q.

=) x² = [ (x +1)² - (x-1)² ] + 60

=) x² = [ ( x² + 1 + 2x) - ( x² + 1 - 2x) ] + 60

=) x² = x² + 1 + 2x - x² - 1 + 2x + 60

=) x² = + 4x + 60

=) x² - 4x - 60 = 0

=) x² - ( 10 - 6 )x - 60 = 0

=) x² - 10x + 6x - 60 = 0

=) x ( x - 10) + 6 ( x - 10) = 0

=) ( x - 10) ( x + 6)

=) [ x = 10 ] and [ x = -6 ].

Since, It's Given that "x" is a Natural Number.

=) [ x = 10 ]

Hence,

The Numbers are;

[x - 1] = [ 10 -1] = 9

x = 10

[ x + 1] = [ 10 + 1] = 11.

Answer 2

Answer:

Three numbers are 9, 10 and 11.

Step-by-step explanation:

Let the numbers be

x, x + 1 and x + 2 (x >0 as x is natural no).

Given :

⇒  (x + 1)² = (x + 2)² - x² + 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 + 7)(x - 9) = 0

x = - 7 or x = 9

Note :

x = - 7 is not possible as x is a natural number

Hence, three numbers are 9, 10 and 11.