<PRE><B>Question 68525</B>
Let x=usual driving speed

Then x+10=increased driving speed

Distance(d)=rate(r) times time(t) or d=rt and t=d/r

Time it takes at usual driving speed=255/x

Time it takes at increased speed=255/(x+10)

24 min=24/60 hr=0.4 hr
Now we are told that increased driving time plus 0.4 hr equals usual driving time.  So our equation to solve is:

 255/(x+10)+0.4=255/x   multiply both sides by x(x+10)  

255(x)(x+10)/(x+10)+(0.4)(x)(x+10)=255/(x+10)(x/x)  clear fractions

255x+0.4x^2+4x=255x+2550  subtract 255x and 2550 from both sides

0.4x^2+4x-2550=0  divide by 0.4
x^2+10x-6375=0----------------------quadratic in standard form

This quadratic can be factored

(x+85)(x-75)=0

x=-85 km/h-----------------discount negative speed
and
x=75 km/h----------------------------usual speed

x+10=75+10=85 km/h-----------------------increased speed


ck

255/75=255/85+0.4
3.4=3.4


Hope this helps----ptaylor</PRE>