Question 28977: The length of a rectangle is 9 meters more than twice the width. If the area of the rectangle is 95 square meters, find the length.
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! The length of a rectangle is 9 meters more than twice the width. If the area of the rectangle is 95 square meters, find the length.
Let the length of the rectangle be L meters
and let the width be B meters.
Lingth is 9 more than twice the width.
L= 2B+9 ----(1)
Area of the rectangle = 95 sqmtrs
That is (LB) = 95 ----(2)
(2B+9)B = 95 (using (1) in (2) (that is substituting for L)
2B^2 +9B -95 = 0 ----(I)
[The product of the square term and the constant term
= (2B^2)X(-95) = -(1X2X5X19)B^2= (19B)X(-10B)
And the middle term (9B) = (19B-10B)]
2B^2 +(19B-10B)-95 = 0
(2B^2 +19B)-10B-95 = 0 (by additive associativity)
B(2B+19)-5(2B+19) = 0
Bp-5p = 0 where p= 2B+19
p(B-5) = 0
p=0 or (B-5)=0
p=0 implies 2B+19 =0 which gives 2B= -19 and therefore B= (-19/2) and this value,we do not take as we get the length and width as negative quaantities
(though according to this B=-19/2,we get length =2B+9 = -19+9 = -10
and length multilplied by width = (-10)X(-19/2) = +95)
B-5 =0 implies B=5
and B=5 in (1) gives L=2B+9 = 2X5 +9 = 10+9= 19
Length =19 meters and width =5 meters.
Verification:Area should be 95 sqmtrs and length X width =19X5 = 95 which is correct. Therefore our values are right
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