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1. The sum of the squares of two consecutive even positive integers is 244.
2. the area of a rectangle is 108cm^2. The length is 3cm greater than the width. Find the width of the rectangle
asked Nov 9, 2013 in ALGEBRA 2 by andrew Scholar

1 Answer

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1) Let " n" is an integer then" 2n " and " 2n+2" be the consecutive even positive integers.

Sum of squares of two consecutive even positive integers is (2n)^2+(2n+2)^2 = 244

4n^2+4n^2+4+8n = 244

8n^2+8n+4-244 = 0

8n^2+8n-240 = 0

Apply division property,divide by 8 to each side.

n^2+n-30 = 0

Do factorization to it.

n^2-5n+6n-30 = 0

n(n-5) +6(n-5) = 0

(n+6)(n-5) = 0

n = -6 or  n = 5

For n = 5

2n = 2(5) = 10

2n+2 = 2(5)+2 = 12

There fore the integers are 10, 12

2) width of rectangle is say b = x

According to given problem length l = x+3

Area = l*b

108 = (x+3)*x

108 = x^2+3x

x^2+3x-108 = 0

x^2+12x-9x-108 = 0

x(x+3)-9(x+3) = 0

(x+3)(x-9) = 0

x =-3, 9

width always in positive form.

For x = 9

width = 9cm

length =9+3 = 12cm

answered Nov 9, 2013 by william Mentor

2) Your answer is correct.

But typo mistake in factorization x2 + 12x - 9x - 108 = 0

 x( x + 12) - 9(x + 12) = 0

(x + 12) (x - 9) = 0

x = -12 and x = 9.

 

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