SOLUTION: 1. the sum of the square of rwo consecutive integers is forty one. find the two integers. 2. the length of a rectangle is 3 in. more than twice its width. its area is 152in^2. f

Algebra ->  Expressions -> SOLUTION: 1. the sum of the square of rwo consecutive integers is forty one. find the two integers. 2. the length of a rectangle is 3 in. more than twice its width. its area is 152in^2. f      Log On


   



Question 174257: 1. the sum of the square of rwo consecutive integers is forty one. find the two integers.
2. the length of a rectangle is 3 in. more than twice its width. its area is 152in^2. find the width of the rectangle
3. find the x-and y- intercepts for 6x-4y=12.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Call the 1st integer n
The next consecutive integer will be n%2B1
n%5E2+%2B+%28n+%2B+1%29%5E2+=+41
n%5E2+%2B+n%5E2+%2B+2n+%2B+1+=+41
2n%5E2+%2B+2n+-+40+=+0
n%5E2+%2B+n+-+20+=+0
Use complete the square method
n%5E2+%2B+n+=+20
n%5E2+%2B+n+%2B+%281%2F2%29%5E2+=+20+%2B+%281%2F2%29%5E2
%28n+%2B+%281%2F2%29%29%5E2+=+%2880+%2B+1%29%2F4
Take the square root of both sides
n+%2B+1%2F2+=+9%2F2
n+=+8%2F2
n+=+4
n+%2B+1+=+5
The numbers are 4 and 5
check answer:
n%5E2+%2B+%28n+%2B+1%29%5E2+=+41
4%5E2+%2B+5%5E2+=+41
16+%2B+25+=+41
41+=+41
OK
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Let w = width
Let l =length
Let A =area
given:
l+=+w+%2B+3
A+=+l%2Aw
A+=+152in2
152+=+l%2Aw
152+=+%28w%2B3%29%2Aw
w%5E2+%2B+3w+-+152+=+0
Use complete the square
w%5E2+%2B+3w+=+152
w%5E2+%2B+3w+%2B+%283%2F2%29%5E2+=+152+%2B+%283%2F2%29%5E2
%28w+%2B+3%2F2%29%5E2+=+%28608+%2B+9%29%2F4
%28w+%2B+3%2F2%29%5E2+=+617%2F4
w+%2B+3%2F2+=+sqrt%28617%29%2F2
w+=+%28sqrt%28617%29+-+3%29%2F2