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

Hey mate,.

Let be X as middle number of the three consecutive numbers.

Then,X+1 and x-1 will be other two numbers.

So,

x²=[(X+1)²-(x-1)²]+60

X²=(x²+2x+1)-(x²-2x-1)+60=4x+60

x²=x²-4x-60

(x-10)(x+6)

So, X can be 10 or -6

.We should consider positive values as X value.

Then , X=10 is the middle number of three consecutive numbers.

And 9 and 11 are other two numbers.

Therefore, 9,10,11are the numbers.

Hope it will help you

✌️sai

Answer 2

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

Hence, three numbers are 9, 10 and 11.