Question 404912
Working together, two people can mow a large lawn in 4 hours.
 One person can do the job alone 1 hour faster than the other person.
 How long does it take each person working alone to mow the lawn?
:
let t = time required by the faster person
then
(t+1) = time required by the other person
:
Let the completed job = 1
:
each man will do a fraction of the work, the two fractions add up to 1
:
{{{4/t}}} + {{{4/((t+1))}}} = 1
:
Multiply by t(t+1) to clear the denominators, results:
4(t+1) + 4t = t(t+1) 
:
4t + 4 + 4t = t^2 + t
:
8t + 4 = t^2 + t
:
0 = t^2 + t - 8t - 4
:
t^2 - 7t - 4 = 0
we have to use the quadratic formula to solve this
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
In this problem x=t, a = 1, b= -7, c = -4
{{{t = (-(-7) +- sqrt(-7^2-4*1*-4 ))/(2*1) }}} 
{{{t = (+7 +- sqrt(49 + 16 ))/2 }}}
{{{t = (+7 +- sqrt(65 ))/2 }}}
Two solutions, only want the positive solution
{{{t = (+7 + 8.06)/2 }}} 
t = {{{15.06/2}}}
t = 7.53 hrs for the faster man to do the joy
and 8.53 hrs for the slower man
:
:
Check solution
4/7.53 + 4/8.53
5.31 + .469  = 1; confirms our solutions of 7.53 and 8.53 hr