document.write( "Question 85375: Sir,\r
\n" ); document.write( "\n" ); document.write( "Pl. help me to solve this problem\r
\n" ); document.write( "\n" ); document.write( "Length of a rectangular field is 23 mtrs more than its Width. If its Area is 420 sq. mtrs, find its length and width.\r
\n" ); document.write( "\n" ); document.write( "I tried like this\r
\n" ); document.write( "\n" ); document.write( "Let Width be = x mtrs
\n" ); document.write( "Then Length is = x+23 mtrs
\n" ); document.write( "(x)+ (x+23) = 420 mtrs
\n" ); document.write( "x2 + 23x = 420 mtrs
\n" ); document.write( "I could not find the value of x. Pl. help me.\r
\n" ); document.write( "\n" ); document.write( "Regards
\n" ); document.write( "

Algebra.Com's Answer #61537 by praseenakos@yahoo.com(507)\"\" \"About 
You can put this solution on YOUR website!
area of the rectangle is = 420
\n" ); document.write( " X^2+23X = 420
\n" ); document.write( " X^2+23X-420 = 0 by factorising 420 we get (35).(-12)
\n" ); document.write( " therefore X^2+35X-12X-420=0
\n" ); document.write( " X(X+35)-12(X+35)=0 or (X+35).(X-12)=0 X-12=0
\n" ); document.write( " X=12
\n" ); document.write( " X+35 =0 X=-35
\n" ); document.write( " the solution can be written as follows
\n" ); document.write( " 1 if X=12 then Y= 12+23=35
\n" ); document.write( " 2 if X=-35 then Y=12-35=-23
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