Hello, Figure_skater123!
You're right . . . We get a quadratic equation.
If Terry had hiked 0.5km/h faster he would have taken 1 hour less to complete a 15km hike.
How fast was he hiking in the first place?
We will use: \(\displaystyle \,\text{Distance }=\text{ Speed }\times\text{ Time}\;\;\Rightarrow\;\;\text{Time }=\:\frac{\text{Distance}}{\text{Speed}}\)
Let \(\displaystyle x\) = Terry's original speed.
He hiked 15 km at \(\displaystyle x\)km/hr. \(\displaystyle \;\)His time was: \(\displaystyle \,\frac{15}{x}\) hours.
If he hiked 15 km at \(\displaystyle x\,+\,0.5\) km/hr, his time would be: \(\displaystyle \,\frac{15}{x\,+\,0.5}\) hours.
This time is one hour less than his actual time: \(\displaystyle \,\frac{15}{x\,+\,0.5}\;=\;\L\frac{15}{x}\,-\,1\)
Multiply through by the LCD, \(\displaystyle x(x\,+\,0.5):\;\;x(x\,+\,0.5)\cdot\left[\frac{15}{x\,+\,0.5}\:=\:\frac{15}{x}\,-\,1\right]\)
\(\displaystyle \;\;\)and we get: \(\displaystyle \,15(x\,+\,0.5)\,-\,x(x\,+\,0.5)\;=\;15x\)
\(\displaystyle \;\;\) which simplifies to: \(\displaystyle \,x^2\,+\,0.5x\,-\,7.5\;=\;0\)
\(\displaystyle \;\;\)which factors: \(\displaystyle \,(x\,-\,2.5)(x\,+\,3)\;=\;0\)
\(\displaystyle \;\;\) and has roots: \(\displaystyle \,x\,=\,2.5,\:-3\)
Since he did not hike backwards, his speed was: \(\displaystyle \,2.5\) km/hr.
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Check
He hiked 15 km at 2.5 km/hr . . . It took him: \(\displaystyle \,\frac{15}{2.5}\,=\,6\) hours.
If he had hiked at \(\displaystyle 2.5\,+\,0.5\:=\:3\) km/hr, it would take him: \(\displaystyle \frac{15}{3}\,=\,5\) hours.
One hour less . . .
check!