Question 205684: A car travels a 750 km journey at average speed of x km/h. If it had increased its speed by 18 km/h, the journey would have been 125 minutes shorter. Form an equation in x and show that it reduces to x²+ 18x= 6480. Solve this equation to find the value of x. Hence find the time taken when the car travels at x km/h.
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! (1) Velocity = Distance/Time,,,v=d/t
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d=750 km,,,,v="x" km/hr,,t= 750/x hr
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(2) d=750,,,v=(x+18),,,,t= { (750/x) -(125/60)} = (750/x)-2.08
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for condition (2),, v=d/t = (x+18) = 750 /{ (750/x) - 2.08},,,,cross multiply
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(x+18){ (750/x)-2.08} =750,,,,,FOIL
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750 -2.08 x +(18*750)/x -37.5 = 750,,,,,simplify and multiply by "x"
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-2.08x^2 +13500-37.5x =0,,,,,,,,divide thru by (-2.08)
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x^2 -6480 +18x = 0,,,,,,or x^2 +18x = 6480,,,,answer
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solving for "x",,,using quadratic formula
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a=(1),,,b=(18),,,c= (-6480)
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x= [ -(18) +/- sqrt{ (18)^2 -4(1)(-6480)}]/2(1)
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x=[ -18 +/- sqrt 26244]/2
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x=[ -18+/- 162]/2
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x=72,,-90(not realistic),,,,,,,,answer
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check by inserting in quadratic problem,,,,both answers ok
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Going back to (1),,,t=d/v = 750/72 = 10.42 hrs,,,,answer
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for (2), t= 750/(72+18)=750/90 = 8.33
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difference of t1-t2 = 10.42-8.33 = 2.08,,,,ok
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