SOLUTION: Cross-country cycling. Erin was traveling across the desert on her bicycle. Before lunch she traveled 60 miles (mi); after lunch she traveled 46 mi. She put in 1 hour more after

Question 191981This question is from textbook
: Cross-country cycling. Erin was traveling across the
desert on her bicycle. Before lunch she traveled 60 miles
(mi); after lunch she traveled 46 mi. She put in 1 hour
more after lunch than before lunch, but her speed was
4 mph slower than before. What was her speed before
lunch and after lunch?
I tried solving this problem by using the d=r(t) equation.
x=rate y=time
before lunch= 60/x=y
after lunch= 48/x-4=y+1
so I substituted 60/x for y
46/x-4=60/x +1
46x=60(x-4) +x(x-4)
46x= 60x-240+x^2-4x
46x=x^2+56x-240
0=x^2+10x-240
I plug this into the quadratic formula, and it doesn't work, any suggestions?
This question is from textbook

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
You're on the right track, you just need to solve the equation

Start with the given equation.

Notice we have a quadratic equation in the form of where , , and

Let's use the quadratic formula to solve for x

Start with the quadratic formula

Plug in , , and

Square to get .

Multiply to get

Rewrite as

Add to to get

Multiply and to get .

Simplify the square root (note: If you need help with simplifying square roots, check out this solver)

Break up the fraction.

Reduce.

or Break up the expression.

So the answers are or

which approximate to or

However since "x" is a speed and a negative speed doesn't make sense, this means that the only answer is

So her speed before lunch was approximately 11.279 mph and her speed after lunch was about 7.279 mph

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