Question

?
Three consecutive positive integers such that square of their sum exceeds
the sum of their squares by 214 are:​

Answer 1

Answer:

this is answer correct options Is D ok

Answer 2

Answer:

the three consecutive integers are  5,6,7

Step-by-step explanation:

let (x-1)  , x  , (x+1) be the 3 consecutive +ve integers

given, [ (x-1) + x + (x+1)]² = (x-1)² + x² + (x+1)² + 214

=> (3x)² = x²-2x+1 + x² + x² + 2x+1 +214

=> 9x² = x² + x² + x² +214 + 1 + 1

=> 9x² = 3x² +216

=> 9x²- 3x² = 216

=> 6x² = 216

=> x² = 216/6 = 36

=> x = √36

=> x = 6

so, the three consecutive integers are ,

(6-1)  , 6  , (6+1)

=> 5,6,7