document.write( "Question 61017: I need help. I am lost on this algebra.\r
\n" ); document.write( "\n" ); document.write( "The area of a rectangle is 119 m2. The base is four less than three times the height. What is the length of the base?
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Algebra.Com's Answer #41945 by checkley71(8403)\"\" \"About 
You can put this solution on YOUR website!
AREA=1/2BASE*HEIGHT
\n" ); document.write( "BASE=3*H-4
\n" ); document.write( "119=1/2(3*H-4)H
\n" ); document.write( "119=1/2(3H-4)H
\n" ); document.write( "119=1/2(3H^2-4H)
\n" ); document.write( "119=1.5H^2-2H
\n" ); document.write( "1.5H^2-2H-119=0 USING THE QUADRATIC EQUATION X=(-B+-SQRT[B^@-4AC])/2A WE GET
\n" ); document.write( "X=(2+-SQRT(4-4*1.5*-119])/2*1.5
\n" ); document.write( "X=(2+-SQRT[4+714])/3
\n" ); document.write( "X=(2+-SQRT[718])/3
\n" ); document.write( "X=(2+-26.8)/3
\n" ); document.write( "X=28.8/3
\n" ); document.write( "X=9.6 SOLUTION. THUS THE HEIGHT IS 9.6M
\n" ); document.write( "119=1/2(B*9.6)
\n" ); document.write( "119=9.6B/2
\n" ); document.write( "9.6B=2*119
\n" ); document.write( "9.6B=238
\n" ); document.write( "B=238/9.6
\n" ); document.write( "B=24.8 SOLUTION & THE LENGTH OF THE BASE.
\n" ); document.write( "PROOF
\n" ); document.write( "119=1/2(9.6*24.8)
\n" ); document.write( "119=1/2(238)
\n" ); document.write( "119=119 \n" ); document.write( "
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