Question

find three cocecutive odd integer ,the sum of whose square is 83

Answer 1

Let three odd consecutive integers be (x+1),(x+3),(x+5)

Accprding to given condition, we have,

(x+1)whole square+(x+3)whole
square+(x+5)whole square=83

x2+1+2x+x2+9+6x+x2+25+10x=83

3x2+35+18x=83

3x2+18x=83-35

3x2+18=48...

divide by 3on given equation, then we have to find that,

x2+6x=16

x2+6x-16

x2+8x-2x-16

x(x+8) -2(x+8)

(x-2) (x+8)

x-2=0 : x+8=0
x=2 : x=-8

the no. r


(x+1),(x+3),(x+5)

=for positive value of x

(2+1),(2+3),(2+5)

****3,5,7 answr



=for negative value of x

(x+1),(x+3),(x+5)

(-8+1),(-8+3),(-8+5)

***-7,-5,-3 answr



hope this helps.....

Answer 2

Let the integers be x, (x+2), (x+4)

Now,

x² +(x+2)² +(x+4)² =83
______________________
By formula (a+b)² =a²+b²+2a
______________________

x²+x²+4+4x+x²+16+8x =83

3x² +20 +12x =83

3x² -83 +20 +12x =0

3x² +12x -63=0

3(x² +4x-21)=0

x² +4x-21 =0

x²+(7-3)x-21 =0

x²+7x-3x-21 =0

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

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

x=3 OR x=-7

Taking positive value,

x=3

Then,

Integers are :-

3

3+2 =5

3+4 =7

I hope this will help you

-by ABHAY