Madison rode her motorcycle 300 miles at a certain average speed. Had she averaged 10 miles per hour more, the trip would have taken 1 hour less. Find the average speed of the motorcycle.

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- Apr 24th 2008, 06:15 PMimbadatmathApplications Involving Quadratic Equations
Madison rode her motorcycle 300 miles at a certain average speed. Had she averaged 10 miles per hour more, the trip would have taken 1 hour less. Find the average speed of the motorcycle.

- Apr 24th 2008, 09:33 PMSoroban
Hello, imbadatmath!

Recall the formula: .$\displaystyle \text{[Distance]} \:=\:\text{[Speed]} \times \text{[Time]} $

We will use the variation: .$\displaystyle T \:=\:\frac{D}{S}$

Quote:

Madison rode her motorcycle 300 miles at a certain average speed.

Had she averaged 10 miles per hour more, the trip would have taken 1 hour less.

Find the average speed of the motorcycle.

She rode 300 miles at $\displaystyle x$ mph.

. . This took her: .$\displaystyle \frac{300}{x}$ hours.

If her speed were $\displaystyle x+10$ mph,

. . it would have taken her: .$\displaystyle \frac{300}{x+10}$ hours.

And this time is one hour less.

There is our equation . . . $\displaystyle \frac{300}{x+10} \;=\;\frac{300}{x} - 1$

Multiply by $\displaystyle x(x+10)\!:\;\;300x \;=\;300(x+10) - x(x+10)$

. . which simplifies to: .$\displaystyle x^2 + 10x - 3000 \:=\:0$

*Go for it!*