<PRE><B>Question 311465</B>
Jim made a 450 mile trip in his car at a certain speed.
 If he had gone 10 mph faster, he could have made the trip in one hour less time.
 Find his speed.
:
Let s = that &quot;certain speed&quot;
then
(s+10) = the faster speed
:
Write a time equation; time = dist/speed
:
Actual time - faster time = 1 hr
{{{450/s}}} - {{{450/(s+10)}}} = 1
multiply by s(s+10), results
450(s+10) - 450s = s(s+10)
:
450s + 4500 - 450s = s^2 + 10s
Arrange as quadratic equation on the right
0 = s^2 + 10s - 4500
Use the quadratic formula to find s
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
In this problem x=s, a=1, b=10 c=-4500
{{{s = (-10 +- sqrt(10^2-4*1*-4500 ))/(2*1) }}}
:
{{{s = (-10 +- sqrt(100 + 18000 ))/(2) }}}
:
{{{s = (-10 +- sqrt(18100 ))/(2) }}}
Two solutions, but we just want positive solution
{{{s = (-10 + 134.536)/(2) }}}
s = {{{124.536/2}}}
s = 62.27 mph his actual speed
:
:
Check solution, find the time at each speed
450/62.27 = 7.2266 hrs
450/72.27 = 6.2266 hrs
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difference: 1 hrs      </PRE>