document.write( "Question 60124This question is from textbook Elementry and Intermediate Algebra
\n" ); document.write( ": If the width of a rectangle is 1 inch less than its length and the diagonal is 1 inch longer than its length, find the length and the width. \n" ); document.write( "
Algebra.Com's Answer #41321 by ptaylor(2198)\"\" \"About 
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Let L=length
\n" ); document.write( "then width (W)=L-1
\n" ); document.write( "and diagonal (D)=L+1
\n" ); document.write( "but the diagonal is actually the hypotenuse of a right triangle that has a length of L and a width of L-1
\n" ); document.write( "and Pythagoras tells us that D=sqrt{L^2+(L-1)^2}
\n" ); document.write( "Thus, the equation that we need to solve is:
\n" ); document.write( "sqrt{L^2+(L-1)^2}=L+1
\n" ); document.write( "Lets go ahead and solve it.
\n" ); document.write( "First, we will square both sides to get rid of the radical which gives us:
\n" ); document.write( "{L^2+(L-1)^2}=L^2+2L+1: Simplifying we get:
\n" ); document.write( "2L^2-2L+1=L^2+2L+1: and further simplifyiing we get:
\n" ); document.write( "L^2-4L=0 and since it's a quadratic, we have two solutions:
\n" ); document.write( "L=0 and L=4; obviously L has to be 4, therefore
\n" ); document.write( "Length (L)=4
\n" ); document.write( "Width (W)=3
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