SOLUTION: Bob and Tom are mowing a yard which is 80' X 40'. They will each mow 1/2 of the yeard. They must mow the yeard in a spiral. How far in does Bob mow before turning it over to Tom? W

Algebra ->  Rate-of-work-word-problems -> SOLUTION: Bob and Tom are mowing a yard which is 80' X 40'. They will each mow 1/2 of the yeard. They must mow the yeard in a spiral. How far in does Bob mow before turning it over to Tom? W      Log On


   

Question 141301: Bob and Tom are mowing a yard which is 80' X 40'. They will each mow 1/2 of the yeard. They must mow the yeard in a spiral. How far in does Bob mow before turning it over to Tom? What equation is used to solve this?
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
80*40/2=(80-x)(40-x)
3200/2=3200-120x+x^2
1600=3200-120x+x^2
x^2-120x+3200-1600=0
x^2-120x+1600=0
using the quadratic equation:x=%28-b%2B-sqrt%28b%5E2-4%2Aa%2Ac%29%29%2F%282%2Aa%29 we get:
x=(120+-sqrt[-120^2-4*1*1600])/2*1
x=(120+-sqrt14,400-6,400])/2
x=(120+-sqrt8,000)/2
x=(120+-89.44)/2
x=(120-89.44)/2
x=30.56/2
x=15.28' is the width of the mowed area that contains one half the total area.