Question 130883
Distance(d)=Rate(r) times Time(t) or d=rt; r=d/t and t=d/r

Let r=speed for first part of trip
Then r-5=speed on second part of trip

Time required for first part of trip= d/r=75/r
Time required for second part of trip=d/r=24/(r-5)

Now we are told that the sum of the above two times equals 5 hours, so:

75/r+24/(r-5)=5  multiply each term by r(r-5)

75(r-5)+24r=5r(r-5)  get rid of parens

75r-375+24r=5r^2-25r  subtract 5r^2 from and add 25r to both sides
75r-375+24r-5r^2+25r=5r^2-5r^2-25r+25r  collect like terms

-5r^2+124r-375=0  multiply each term by -1

5r^2-124r+375=0  quadratic in standard form.  Solve using quadratic formula
{{{r = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{r = (124 +- sqrt( (-124)^2-4*5*(-375) ))/(2*5) }}}  
{{{r = (124 +- sqrt(7876 ))/(10) }}}  
{{{r = (124 +-88.75)/(10) }}} 
{{{r = (124 +88.75 )/(10) }}} 
{{{r = 21.275}}} mph------speed on first part of trip
{{{r-5=21.275-5=16.275}}} mph---speed on second part of trip


{{{r = (124 -88.75 )/(10) }}} 
{{{r = 3.525}}} mph--------------speed on second part of trip (NO!!!)
r-5=3.535-5=NEGATIVE VALUE  NO GOOD!!!!


CK

75/21.275 + 24/16.275=5
3.525+1.474=5
4.999999--~~~~5


Hope this helps---ptaylor