Question

If the product of three consecutive positive integers is equal to their sum,then what would be the sum of their squares?

OPTIONS:

a.9
b.14
c.16
d.24

Answer 1

Hey mate..
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Let , the three consecutive numbers be x , ( x + 1 ) and ( x + 2 ) respectively.


Given,

The product of three consecutive positive integers is equal to their sum.

So,

x (x+1)(x+2) = x+(x+1)+(x+2)

=> x { x ( x+2 ) +1 ( x+2 ) } = x+x+1+x+2

=> x (x^2 + 2x + x + 2) = x+x+1+x+2

=> x (x^2 + 3x + 2) = x+x+1+x+2

=> x^3 + 3x^2 + 2x = 3x + 3

=> x^3 + 3x^3 - x - 3 = 0

=> x^2 (x + 3) - 1 (x + 3) = 0

=> ( x + 3 ) (x^2 - 1) = 0

=> (x+3) (x-1) (x+1) = 0

Either,

(x + 3) = 0

i.e. x = -3

Or,

(x - 1) = 0

I.e. x = 1

Or,

(x - 1) = 0

i.e. x = 1

Here,

There is only one positive integer i.e. 1

So,

The three consecutive positive integers are 1 , 2 and 3 respectively.

And the Sum of the squares are:--

(1)^2 + (2)^2 + (3)^2

= 1 + 4 + 9

= 14 ( Ans:// )

So, B) 14 is the correct option .

#racks

Answer 2

Answer:

14

Step-by-step explanation:

let the three consecutive positive integer be x , x+1 , x+2

x * (x+1) * (x+2) =x + x+1 + x+2

x ( x(x+2) +1 (x+2)) = 3x + 3

x (  x^2 + 2x +1x +2 ) = 3x +3

x ( x^2 + 3x + 2 ) = 3x + 3

x^3 + 3x^2 + 2x = 3x + 3

x^3 + 3x^2 + 2x - 3x -3 = 0

x^3 + 3x^2 -x -3 = 0

x^2 ( x + 3 ) -1 (x + 3) =0

(x + 3 )   ( x^ 2 - 1 )

(x + 3 )   (  x + 1  )    ( x - 1 )

x + 3 = 0          x + 1 = 0             x - 1  = 0

x = -3                x = -1                 x = 1

The only one that's positive is  x= 1,

So the three consecutive integers are  1 , 2 , 3

The sum of their squares = 1^2+2^2+3^2

                                           = 1+4+9

                                           = 14

#SPJ2