Question

Find three consecutive odd integers the sum of whose squares is 83​

Answer 1

Step-by-step explanation:

let the three consecutive Odd integer

X, X+2,X+4

according to question

x^2+(X+2)^2+(X+4)^2=83

x^2 +x^2+4+4x+x^2+16+8x=83

3x^2+20+12x=83

3x^2+12x-63=0

x^2+4x-21=0

x^2+7x-3x-21=0

X(X+7)-3(X+7)=0

(x-3)(X+7)=0

x=3 or x= -7

hence the number is

3,5,7

Answer 2

Answer:

3, 5, 7 or -7, -5, -3

Step-by-step explanation:

Assuming the smallest number is x,

smallest number = x

middle number = x+2

Largest number = x+4

By summing the squares,

$x^{2}$ + $(x+2)^{2}$ +$(x+4)^{2}$ = 83

$3x^{2}$ + 12$x$ + 20 = 83

$3x^{2}$ + 12$x$ = 63

$3x^{2}$ + 12$x$ - 63 = 0

Divide by 3,

$x^{2}$ + $4x$ -21 = 0

$(x-3)(x+7)$ = 0

$x = 3$ or -7

Substitute x = 3 into the integers,

smallest number = 3

middle number = 3+2

                          = 5

Largest number = 3+4

                           = 7

Substitute x = -7 into the integers,

smallest number = -7

middle number = - 7 + 2

                          = - 5

Largest number = - 7 + 4

                           = - 3