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2006-12-10 03:45:50 · 10 answers · asked by oldmansaggins4u 1 in Science & Mathematics Mathematics

10 answers

You can just figure this out by trial and error. They have to average 77/3 = 25, so it's going to be right around 5.

if we use 4, 5, 6, we get 16, 25, 36, which sum to 77 :)

2006-12-10 03:48:28 · answer #1 · answered by moto 3 · 0 2

2 solutions : 4, 5, 6 or -6, -5, -4 sixteen + 25 + 36 = 77 i used trial and mistake however the not undemanding way is: n^2 + (n + a million)^2 + (n + 2)^2 = 77 n^2 + n^2 + 2n + a million + n^2 + 4n + 4 = 77 3*n^2 + 6n + 5 = 77 3*n^2 + 6n - seventy two = 0 then use quad. style. to locate n = -6 or 4 the different 2 numbers interior the triplet may well be n + a million and n + 2.

2016-12-11 06:15:37 · answer #2 · answered by jeniffer 4 · 0 0

Let the the squares of three consecutive intergers be
x^2 , (x + 1)^2 and (x + 2)^2.

x^2+(x+1)^2+(x+2)^2=77
x^2+(x^2+2x+1)+(x^2+4x+4)=77
3x^2+6x+5=77
3x^2+6x-72 = 0
Devide each term by 3,
x^2+2x-24=0
(x+6)(x-4)=0
x=-6, 4
When x=-6, x+1=-5, x+2=-4
When x=4, x+1=5, x+2=6

Hence, the three consecutive intergers are -6,-5,-4 or 4,5,6.

2006-12-10 05:20:25 · answer #3 · answered by Ranna Renni 2 · 0 0

4^2 = 16 5^2 = 25 6^2 =36

16 + 25 + 36 = 77

2006-12-10 03:49:15 · answer #4 · answered by Anonymous · 0 1

It's 4,5,6 and also -4,-5,-6, since when you square the negatives, it becomes positive
(x)^2 + (x+1)^2 + (X+2)^2= 77
You distribute then solve the quadratic and get -6 and 4 for x
then add 1 and 2 to the x values to solve

2006-12-10 03:50:03 · answer #5 · answered by Panky1414 2 · 1 0

Consecutive integers can always given by:
x-1
x
x+1

So: (x-1)^2 + x^2 + (x+1)^2 = 77

x^2 -2x + 1 + x^2 + x^2 + 2x + 1 = 77

Gather like terms:
3x^2 + 2 = 77
3x^2 = 75
x^2=25

So, x= 5 and -5 {NOTE: -5 is a perfectly good integer!}

Thus, the answer is : 4,5,6 and -6, -5,-4.

2006-12-10 03:54:28 · answer #6 · answered by Jerry P 6 · 1 0

4, 5 and 6

4 squared is 16
5 squared is 25
6 squared is 36

Equals 77

2006-12-10 03:48:35 · answer #7 · answered by Mark W 2 · 0 1

x^2 + (x + 1)^2 + (x + 2)^2 = 77
x^2 + ((x + 1)(x + 1)) + ((x + 2)(x + 2)) = 77
x^2 + (x^2 + x + x + 1) + (x^2 + 2x + 2x + 4) = 77
x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) = 77
x^2 + x^2 + 2x + 1 + x^2 + 4x + 4 = 77
3x^2 + 6x + 5 = 77
3x^2 + 6x - 72 = 0
3(x^2 + 2x - 24) = 0
x^2 + 2x - 24 = 0
(x + 6)(x - 4) = 0
x = -6 or 4

ANS : 4,5,6 or -4,-5,-6

2006-12-10 04:37:29 · answer #8 · answered by Sherman81 6 · 0 0

n^2 + (n+1)^2 + (n+2)^2 = 77
solve for n

2006-12-10 03:48:58 · answer #9 · answered by Anonymous · 0 1

4,5,6
16,25,36=77

2006-12-10 03:55:39 · answer #10 · answered by moha 1 · 0 1

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