document.write( "Question 90301This question is from textbook Beginning Al
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document.write( ": The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm, what are the dimensions (the length and the width of the rectangle?\r
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document.write( "x^2 + (x + 1)^2 = 4^2
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document.write( "x^2+ x^2 + 2x + 1 = 16
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document.write( "2x^2 + 2x + 1 -16 = 0
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document.write( "2x^2 + 2x -15 = 0
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document.write( "x^2 + x -15/2 = 0\r
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document.write( "I think the above is correct, but now I don't know what to do...
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Algebra.Com's Answer #65552 by ptaylor(2198)![]() ![]() You can put this solution on YOUR website! YOU WERE ALMOST THERE!!!!!!!!!!!!!!! \n" ); document.write( "Let x=width \n" ); document.write( "Then x+1=length\r \n" ); document.write( "\n" ); document.write( "x^2+(x+1)^2=4^2 get rid of parens\r \n" ); document.write( "\n" ); document.write( "x^2+x^2+2x+1=16 \n" ); document.write( "2x^2+2x+1-16=0\r \n" ); document.write( "\n" ); document.write( "2x^2+2x-15=0 quadratic in standard form. We can solve using the quadratic formula\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "We'll not worry about the negative value for x, since lengths and widths are positive \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "CK\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "16.001~~16\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |