Sum of consecutive integers plus word problem?

Question

Find two consecutive "positive" integers such that the sum of thier squares is 85. Setting up the equation with a variable.

Also this word problem is confusing:

A garden area is 30ft long and 20 ft wide. A path of uniform width is set around the edge. IF the remaining garden area is 400ft^2, what is the width of the path?

Thanks for the help !

Answer 1

n and m such that
m=n+1

n^2 + m^2 = 85
n^2 + (n+1)^2 = 85
n^2 + n^2 + 2n + 1 = 85
2n^2 + 2n -84 = 0

Solve for n

n = 6, m = 7

Garden area = 30*20 = 600 ft^2
now you have a path of width x which makes it 400 so

(30-2x)(20-2x) = 400
4x^2 -100x + 200 = 0

Solve for x

x=2.192236 ft

Answer 2

1) Let the consecutive positive integers be x and x + 1.
Thus, x^2 + (x + 1)^2 = 85

Expanding: x^2 + x^2 + 2x + 1 = 85
Summing and rearranging: 2x^2 + 2x - 84 = 0
Dividing through by 2: x^2 + x - 42 = 0
Factorising: (x - 6)(x + 7) = 0
Equating each term to zero: x = 6 or -7
But x must be positive, therefore x = 6.

2) Let w = width of path
I think the problem is saying that the path is set inside the
existing garden area, so that the total area of the now, small
garden, plus the area of the path is 30 ft * 20 ft = 600 ft^2.

This means that the path area = 600 - 400 = 200 ft^2

Now the length of the new garden area will be 30 - 2w
and the width will be 20 - 2w.
Thus, (30 - 2w)(20 - 2w) = 200

Expanding: 600 - 60w - 40w + 4w^2 = 200
Summing and rearranging: 4w^2 - 100w + 400 = 0
Dividing through by 4: w^2 - 25w + 100 = 0
Factorising: (w - 20)(w - 5) = 0
Equating each term to zero: w = 20 or 5.
But if w = 20, then both length and width will be negative.
So w = 5 ft.
Therefore, length of garden = 30 - 2*5 = 20 ft
and the width = 20 - 2*5 = 10 ft.

Answer 3

If x is the first positive integer, the next consecutive integer is x+1, so the sum of two consecutive positive integers is written x^2 + (x+1) ^2.

x^2+(x+1)^2 = 85 => 2x^2 + 2x = 84 => x(x+1) = 42 => x = 6

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



The garden area has a total area of 600 ft^2. So the total area of the path is 200 ft^2. Let the width of the path be x, then the total length of the path is 2*(30-x)+2*(20-x), so the area becomes 100x - 4x^2 = 200. This is a basic quadratic, set to 0 leaves -4x^2 + 100x - 200 = 0. The quadratic formula can be used =>

(-b +- sqrt(b^2-4ac)) / 2a; Where a = -4, b = 100, c = -200.

(-100 +- sqrt(10000 - 3200))/-8 => (-100 +- sqrt(6800))/-8 = (25-5sqrt(17))/2 ~ 2.19.

The larger value isn't possible since the garden area is only 20 feet wide, so the path has width x = (25-5*sqrt(17))/2 ~ 2.192236...

Answer 4

As an equation:
x^2 + (x+1)^2 = 85
so...
2x^2 + +2x + 1 = 85

x^2 + x = 42

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

so...
x = 6

Area = Width * Length
Garden area = 30 ft * 20 ft = 600 ft^2
Garden perimeter = 2 * 30 ft + 2 * 20 ft = 100 ft
The perimeter around the garden will be the length of the path, so...
Garden perimeter = Path Length

Remaining Area (RA) = 400 ft^2
RA = Path Width (PW) * Path Length(PL)
400 ft^2 = PW * 100 ft

PW = (400 ft^2)/100 ft
Path Width = 4 ft

Answer 5

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

n^2+n-42=0
(n-6)(n+7)=0
therefore n=6 and 7

the garden area is 300x200=600 and the area of the path is 600-400=200
let width be x
2(20x)+2[(30-x)x]=200
40x+2[(30x-x^2)=200
x^2-100x+200=0

Answer 6

the anwers to teh first one is 6 and 7
x^2 + (x+1)^2 = 85 (maybe, i solved 6 and 7 in my head)

Answer 7

a^2 + b^2 = 85
b = a+1

so a = 6