Question

the sum of the square of three consecutive positive integers is 50.find the integers

Answer 1

Let the first number be x
second be x + 1
third be x + 2


▪ The sum of squares is 50

x² + ( x + 1 )² + ( x + 2 )² = 50

x² + x² + 1 + 2x + x² + 4 + 4x = 50

3x² + 5 - 50 + 6x = 0

3x² - 45 + 6x = 0

x² - 15 + 2x = 0

x² + 2x - 15 = 0

x² + 5x - 3x - 15 = 0

x ( x + 5 ) - 3 ( x + 5 ) = 0

( x - 3 ) ( x + 5 ) = 0


* ( x - 3 ) = 0

x = 3


* ( x + 5 ) = 0

x = -5 ( neglected , because it is given that the integer are positive )


So,

First number x = 3

second number x + 1

= 4

Third number x + 2

= 5


So, the numbers are 3 , 4 and 5

Answer 2

Let Integers be x,y and z
We are given that Integers are consecutive so we can also say that integers are x,x+1,x+2

We are given, x²+y²+z²= 50
                 or, x²+(x+1)²+(x+2)²= 50
                      x²+x²+1+2x+x²+4+4x= 50    {∵ (a+b)²=a²+b²+2ab}
 
                      3x²+6x+5= 50
                      3x²+6x+5-50= 0
                      3x²+6x-45= 0
                      3x²+15x-9x-45= 0           [By Splitting Middle Term]
                      3x(x+5) -9(x+5)= 0
                         
                          so, x+5=0 or 3x-9=0
                          so, x=-5 or x= 9/3 = 3

               But Given that Integers are positive so consider x=3
so ,Numbers are x,x+1 and x+2= 3,4,5