Algebra - Wind problem

Hello, SwissArmelle!

I'll do it the "standard" way", but we'll get a tricky equation . . .

Quote:

During a 3000-mile flight, a plane encountered a strong tail wind
that increased its speed by 100 mi/hour.
This increase of speed shortened the flying time by one hour.
Find the speed of the plane relative to the ground.

We will use: . $\text{Distance} \:=\: \text{Speed} \times \text{Time}\quad\Rightarrow\quad t \:=\:\frac{d}{s}$

Let $S$ = speed of the plane (ground speed).

Normally, it would fly the 3000 miles at $S$ mph.
. . It would take: . $\frac{3000}{S}$ hours.

With the tailwind, its speed is $S + 100$ mph.
. . To fly 3000 miles, it took only: . $\frac{3000}{S + 100}$ hours.
And this is one hour less than its normal time.

There is our equation . . . . $\frac{3000}{S + 100} \:=\:\frac{3000}{S} - 1$

Multiply by $S(S+100)\!:\;\;3000S \:=\:3000(S + 100) -S(S + 100)$

This simplifies to the quadratic: . $S^2 + 100S - 30,000 \:=\:0$

. . which factors: . $(S - 500)(S + 600) \:=\:0$

. . and has roots: . $S \:=\:500,\:-600$

Therefore, the plane's ground speed is: $\boxed{500\:mph}$