) find two consecutive positive integers such that the sum of their square is 85.?

Answer 1

If the sum of two consecutive numbers' squares is 85, the average of the squares is half that, or 42.5.

So you pick the two numbers (6, 7) whose squares (36, 49) are on either side of 42.5

Answer 2

Let x = first integer then (x + 1) = 2nd integer x² + (x + 1)² = 85 x² + x² + 2x + 1 = 85 2x² + 2x - 84 = 0 Divide thru by 2 x² + x - 42 = 0 factor (x + 7)(x - 6) = 0 Discard trivial negative solution x = 6 x + 1 = 7 The two integers are 6 and 7 .

Answer 3

two consecutive integers: x, x+1

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

x^2 + x^2 + 2x + 1 =85

2x^2 + 2x + 1 = 85

2x^2 + 2x = 85 - 1 = 84

2(x^2 + x) = 84

x^2 + x = 84/2 = 42

x(x + 1) =42 you should be able to see right away that the consecutive positive integers must be 6 and 7, but in case you can't, x(x + 1) represent the two factors of 42 which satisfy the condition (notice that they also verify that the answer can only be two consecutive positive integers).

factor 42: 1*42, 2*21, 3*14, 6*7

which of these combinations represents two consecutive positive integers? 6*7

Answer 4

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

x = 6 or x = -7
Since in your problem, x had to be positive.. then x = 6

If x = 6, then the next consecutive integer is
x + 1 = 7

So 6 and 7 are your two consecutive integers

Answer 5

6 and 7

To solve this, let's the define the following terms:

x = "the first integer"
x + 1 = "the second integer"

x² + (x + 1)² = 85
==> FOIL the second term
x² + x ² + 2x + 1 = 85
==> combine like terms
2x² + 2x + 1 = 85
==> subtract 85 from both sides
2x² + 2x - 84 = 0
==> divide everything by 2
x² + x + - 42 = 0
==> factor
(x - 6)(x - 7) = 0
==> this yields two solutions, which are your 2 numbers
x = 6 or x = 7

Answer 6

given:
x^2+y^2=85 (sum of their square is 85)
y=x+1 (two consecutive positive integers x,y)

substitute y

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

take the roots,
we have x=-7 and x=6
we take x=6 since it is a positive integer.
then substitute x=6 into y.
we have y=x+1,,,
y=6+1
y=7

therefore the answer is 6 and 7

Answer 7

Let x = 1st integer, x + 1 = 2nd integer.

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

Answer: 1st integer = 6, 2nd integer = 7

Proof:
6^2 + 7^2 = 85
36 + 49 = 85
85 = 85

Answer 8

5 and 6

Answer 9

6 and 7

6^2 + 7^2 = 36 + 49 = 85

Answer 10

6^2 + 7^2 = 85