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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.
\n" ); document.write( "Let the length of the rectangle be L meters
\n" ); document.write( "and let the width be B meters.
\n" ); document.write( "Lingth is 9 more than twice the width.
\n" ); document.write( "L= 2B+9 ----(1)
\n" ); document.write( "Area of the rectangle = 95 sqmtrs
\n" ); document.write( "That is (LB) = 95 ----(2)
\n" ); document.write( "(2B+9)B = 95 (using (1) in (2) (that is substituting for L)
\n" ); document.write( "2B^2 +9B -95 = 0 ----(I)
\n" ); document.write( "[The product of the square term and the constant term
\n" ); document.write( "= (2B^2)X(-95) = -(1X2X5X19)B^2= (19B)X(-10B)
\n" ); document.write( "And the middle term (9B) = (19B-10B)]
\n" ); document.write( "2B^2 +(19B-10B)-95 = 0
\n" ); document.write( "(2B^2 +19B)-10B-95 = 0 (by additive associativity)
\n" ); document.write( "B(2B+19)-5(2B+19) = 0
\n" ); document.write( "Bp-5p = 0 where p= 2B+19
\n" ); document.write( "p(B-5) = 0
\n" ); document.write( "p=0 or (B-5)=0
\n" ); document.write( "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
\n" ); document.write( "(though according to this B=-19/2,we get length =2B+9 = -19+9 = -10
\n" ); document.write( "and length multilplied by width = (-10)X(-19/2) = +95)
\n" ); document.write( "B-5 =0 implies B=5
\n" ); document.write( "and B=5 in (1) gives L=2B+9 = 2X5 +9 = 10+9= 19
\n" ); document.write( "Length =19 meters and width =5 meters.
\n" ); document.write( "Verification:Area should be 95 sqmtrs and length X width =19X5 = 95 which is correct. Therefore our values are right
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