Help with Quadratic formula word problem

[Figure_skater123]

My question is:
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?

I think I have to use the Quadratic formula, to solve the question, but i don't even know where to start.

 

[soroban]

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!

 

[Denis]

Have you TRIED, starting with speed = distance / time?

speed = x, time = t

x = 15 / t [1]
x + 1/2 = 15 / (t - 1) [2]

[1]: 15 = xt
[2]: 15 = (x + 1/2)(t - 1)

Carry on; you will get a quadratic.

 

[Figure_skater123]

Hello, again. Thank you for your help. I sort of forgot I posted this, thats why it has taken me so long to reply.