integer problem?

Question

The sum of the squares of two consecutive positive integers is 85. Find the integers.

I know the answer is 6, 7 but how do I get to that answer? if someone can please explain.

Answer 1

OK - let x be the smaller number.

So x^2 + (x+1)^2 = 85. Now let's simplify.

x^2 + x^2 + 2x + 1 = 85
2x^2 + 2x = 84
x^2 +x = 42
x^2 +x -42 = 0
(x +7)(x-6) = 0

So the answers are -7 and 6. But since we are looking for positive integers, -7 is out and x (the smaller number ) is 6.

That means x+1 =7. So the number are 6 ,7.

Check
6^2 = 7^2 = 85?
36 + 49 =85?
85 = 85 YES!!

Hope this helps.

Answer 2

okay so let's say that our first number is x. then the next integer is x+1 right?

now the sum of the squares can be written like this
(x)^2+(x+1)^2
and we know this is equal to 85 so
(x)^2+(x+1)^2=85

so
(x)^2+(x+1)(x+1)=85

so after foiling, or whatever method you use you get:
(x)^2+(x)^2+2x+1=85

combine the like terms and you get
2(x)^2+2x+1=85
subtract 85 from both sides and you get

2(x)^2+2x-84=0
all the terms on the left side are divisible by 2 so divide both sides by 2 to make it easier on you.

you get
(x)^2+x-42=0

now factoring you get
(x+7)(x-6)=0
so x can be -7 or 6. but your teacher said it had to be two consecutive POSITIVE integers. so -7 is out of the question.

so the first number is 6.
the next number is x+1, 6+1 =7

Answer 3

Let the consecutive numbers be N and (N+1) .

Sum of squares means

N^2 + (N+1)^2 = 85, expand

N^2 + N^2 + 2N + 1 = 85, gather

2N^2 + 2N -84 = 0, divide by 2

N^2 + N - 42 = 0, factor

(N +7)(N -6) = 0, so

N + 7 = 0 | N - 6 = 0
N= -7 | N = 6

We want the numbers positive, so N = -7 is out.

So we have N=6, N+1 = 7
Done.

Answer 4

Okay, let us call the smaller of the two consecutive integers X, then the larger one will be X+1.

The sum of their squares is X^2 + (X+1)^2, and we want the value of X which makes that come to 85.

X^2 + (X^2 + 2X + 1) = 85, subtract 1 from each side:
2X^2 + 2X = 84, double it:
4X^2 + 4X = 168, add 1 to each side:
4X^2 + 4X + 1 = 169, express as perfect squares:
(2X + 1)^2 = 13^2, take square root of both sides:
2X + 1 = 13, 2X = 12, X = 6

The smaller number is 6, and the larger is 7.

Answer 5

Let the two integers be x and x + 1.

x² + (x + 1)² = 85
x² + x² + 2x + 1 - 85 = 0
2x² + 2x - 84 = 0
x² + x - 42 = 0
(x + 7)(x - 6) = 0
x = 6
since the integers are positive

x + 1 = 7

The two integers are 6 and 7.

Answer 6

Let x be the first number, so the next one will be (x+1).

x^2+(x+1)^2=85
x^2+x^2+2x+1=85
2x^2+2x+1=85

The roots of this quadratic are x=6 and x=-7, but you are looking for positive integers, so the only acceptable solution is x=6.

Since x=6 then x+1=7

Answer 7

let x,y be the unknown,
since it's consecutive,
y=x+1

x^2 + y^2 = 85
x^2 + (x+1)^2 = 85
expand and simplify the eqn,
2x^2 + 2x - 84 = 0
x^2 + x - 42 = 0
(x-7)(x+6)=0
x=7 or -6
since it says +ve interger,
x=7
when x=7
7^2 + y^2 = 85
y^2 = 85 - 49
y = sqrt 36
y = 6

Answer 8

Let the numbers be x and (x+1)

x² + (x+1)² = 85
Expand:
x² + x² + 2x + 1 = 85
2x² + 2x - 84 = 0
x² + x - 42 = 0
x² + 7x - 6x - 42 = 0
x(x+7)-6(x+7) = 0
(x+7)(x-6) = 0
x = {-7, 6}

The number MUST be positive
x = 6
x+1 = 7

The numbers are 6 and 7

Answer 9

x² + (x + 1)² = 85
x² + x² + 2x + 1 = 85
2x² + 2x - 84 = 0
x² + x - 42 = 0
(x + 7)(x - 6) = 0
x = - 7 , x = 6
Integers are 6 and 7

Answer 10

x^2 +(x+1)^2 =85
x^2+ x^2 +2x+1 =85
2x^2 +2x -84 =0
2(x^2 +x-42)
(x+7)(x-6)