Question 49045
OK PLEASE SEE BELOW

Question 49045: Please Explain: 
A rectangular lawn is 24 feet wide by 32 feet long. A sidewalk will be built along the inside edges of all four sides. The remaining lawn will have an area of 425 square feet. How wide will the walk be?: Please Explain: 
LET THE WIDTH OF PATH =X
L=32'............W=24'
LENGTH OF INNER RECTANGLE=L'=L-2X=32-2X...SINCE ON BOTH SIDES PATH TAKES X+X FEET LEAVING 32-2X FOR THE GARDEN PROPER.
WIDTH OF INNER RECTANGLE=W'=W-2X=24-2X
HENCE REMAINING GARDEN AREA=L'*W'=(32-2X)(24-2X)=425
768-64X-48X+4X^2=425
4X^2-112X+343=0..THIS IS A QUADRATIC.OF TYPE AX^2+BX+C=0
ITS ROOTS ARE
X=[-B+ OR - SQRT.{B^2-4AC}]/2A
WE HAVE HERE A=4..B=-112...C=343...SO
X=[112-SQRT(112^2-4*4*343)]/8=  3 '



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This is a word problem that I'm sure uses the quadratic formula because this is what the chapter is about! The formula seems simple enough, but I don't know how to set this up.
A garden area is 30ft. long (L)and 20ft. wide(W). A path of uniform width is set around the edge. If the remaining garden are is 400ft^2, what is the width of the path?
LET THE WIDTH OF PATH =X
L=30'............W=20'
LENGTH OF INNER RECTANGLE=L'=L-2X=30-2X
WIDTH OF INNER RECTANGLE=W'=W-2X=20-2X
HENCE REMAINING GARDEN AREA=L'*W'=(30-2X)(20-2X)=400
600-60X-40X+4X^2=400
4X^2-100X+200=0
X^2-25X+50=0
X=[25-SQRT(25^2-4*50)]/2=2.19'