document.write( "Question 28932: A 150 mile trip is driven at R miles per hour. The same trip would take 2 hours less if you increase the speed by 20 mph. What is R? I have tried R=150/t
\n" ); document.write( "R=150/t-2 (150/t+2=150/t-2).t(t-1)= 150t-150t+2t2-2t= t2-t-150=0 = (t+2)(t+75)and that is as far as I get because I know that that is not right.
\n" ); document.write( "Thank you for your help. These always put me in a tail spin. Cheryl Vogel
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Algebra.Com's Answer #15806 by Paul(988)\"\" \"About 
You can put this solution on YOUR website!
let the normal speed be
\n" ); document.write( "If increased by 20 the speed will be x+20
\n" ); document.write( "Subtract the increase speed by the normal speed respect to the distance = time.\r
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\n" ); document.write( "\n" ); document.write( "Equation:
\n" ); document.write( "\"150%2Fx-150%2F%28x%2B20%29=2\"
\n" ); document.write( "\"150%28%28x%29-%28x%2B20%29%29=2%28%28x%29%28x%2B20%29%29\"
\n" ); document.write( "\"150%2820%29=2%28x%5E2%2B20x%29\"
\n" ); document.write( "\"3000=2x%5E2%2B40x\"
\n" ); document.write( "\"2x%5E2%2B40x-3000\"
\n" ); document.write( "\"x%5E2%2B20x-1500=0\"
\n" ); document.write( "a=1, b=20, c=-1500\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"
\n" ); document.write( "\"x=%28-20%2B-sqrt%2820%5E2-4%2A1%2A-1500%29%29%2F%282%2A1%29\"
\n" ); document.write( "Simplfy that and you get 2 solutions:\r
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\n" ); document.write( "\n" ); document.write( "x=-50 and x=30\r
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\n" ); document.write( "\n" ); document.write( "Remove the negative\r
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\n" ); document.write( "\n" ); document.write( "Hence, the normal speed (R) is 30mph.
\n" ); document.write( "Paul.\r
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