SOLUTION: Two men start at the same time from the same place and travel along roads that are at right angles to each other. One man travels 4mi/hr faster than the other, and at the end of tw

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Question 128706: Two men start at the same time from the same place and travel along roads that are at right angles to each other. One man travels 4mi/hr faster than the other, and at the end of two hours they are 40mi apart. Determine their rates of travel.
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
X^2+(X+4)^2=40^2
X^2+X^2+8X+16=1600
2X^2+8X+16-1600=0
2X^2+8X-1584=0
2(X^2+4X-792)=0
USING THE QUADRATIC EQUATION:x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+WE GET:
X=(-4+-SQRT[4^2-4*1*-792])/2*1
X=(-4+-SQRT[16+3168])/2
X=(-4+-SQRT3184)/2
X=(-4+-56.427)/2
X=(-4+56.427)/2
X=52.426/2
X=26.213 MPH. ANSWER FOR THE SLOWER TRAVELER.
26.213+4=30.213 MPH. FOR THE FASTER TRAVELER.
PROOF:
26.213^2+30.213^2=40^2
687.12+912.83=1600
1599.95~1600