Question 82011
distance(d)=rate(r) times time(t) or d=rt;t=d/r and r=d/t

Let r=speed of the wind

time airplane traveled against the wind=716/(224-r)
time airplane traveled with the wind =1904/(224+r)

Now we are told that the sum of these times is 12 hrs.  So:

716/(224-r)+1904/(224+r)=12  multiply each term by (224-r)(224+r) to get rid of fractions:

716(224-r)(224+r)/(224-r)+1904(224-r)(224+r)/(224+r)=12(224-r)(224+r) simplify:

716(224+r)+1904(224-r)=12((224-r)(224+r)  get rid of parens

160384+716r+426496-1904r=602112-12r^2  simplify

586880-1188r=602112-12r^2  subtract 602112 and add 12r^2 to both sides

12r^2-1188r-15232=0  divide each term by 4

3r^2-297r-3808=0  quadratic in standard form.  Solve using quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{x = (297 +- sqrt( 297^2-4*3*(-3808) ))/(2*3) }}}

{{{x = (297 +- sqrt(133905))/(6) }}}

{{{x = (297 +- 365.930)/(6) }}}

{{{x=110.4883}}} mph----------------------speed of the wind

discount the negative value of x

CK
716/(224-110.4883)+1904/(224+110.4883)=12
716/(113.5116)+1904/(334.4883)=12
6.3077+5.6923=12
~12=12

----a nasty problem with respect to manipulating the terms---there may be an easier way, but I don't see it.  It's good that we have calculators!


Hope this helps----ptaylor