document.write( "Question 871931: Paul drove to the city, 200 miles away, and returned home at the speed which was 10 mph faster than on his trip to the city. If the trip to the city was one hour longer than his return trip, what was Paul's speed from home to the city? \n" ); document.write( "
Algebra.Com's Answer #525800 by mananth(16946)\"\" \"About 
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going speed x
\n" ); document.write( "returning speed x + 10
\n" ); document.write( "
\n" ); document.write( " Distance = same 200 miles 1
\n" ); document.write( " original time – time with increased speed = 1
\n" ); document.write( " t=d/t
\n" ); document.write( "200 / x - 200 / ( x + 10 ) = 1
\n" ); document.write( "LCD= x ( x + 10 )
\n" ); document.write( "multiply by LCD
\n" ); document.write( "200 ( x + 10 ) - 200 x = 1 x ( x + 10 )
\n" ); document.write( "200 x + 2000 - 200 x = 1 X^2 + 10 x
\n" ); document.write( "2000 = 1 X^2 + 10 x
\n" ); document.write( "1 X^2 + 10 x - -2000 = 0
\n" ); document.write( "Find roots of the quadratic equation
\n" ); document.write( "a= 1 b= 10 c= -2000
\n" ); document.write( "
\n" ); document.write( "x1= ( -10 + sqrt( 100 - 8000 )) / 2
\n" ); document.write( "x1=( -10 + 90 )/ 2
\n" ); document.write( "x1= 40
\n" ); document.write( "
\n" ); document.write( "x2= ( -7 - sqrt( 49 - 20 ) / 2
\n" ); document.write( "x2=( -10 - 90 )/ 2
\n" ); document.write( "x2= -50
\n" ); document.write( "ignore negative\r
\n" ); document.write( "\n" ); document.write( "going speed = 40 mph
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