# Identifying Quadratic equations from collected information

A girl can row her boat at $5 km/h$ in still water. If she takes $1$ hour more to row the boat $5.2 km$ upstream then to return downstream, find the speed of the stream.

What I had done so far:

Let,

• Velocity of water be $x km/h$

We know,

Velocity of boat is $5 km/h$

Velocity of boat in upstream = $(5-x) km/h$

Velocity of boat in downstream = $(5+x) km/h$

Distance = $5.2 km$

Time taken to cover distance in upstream = $\frac{5.2}{5-x} hours$

Time taken in downstream = $\frac{5.2}{5+x} hours$

Now how can I create a quadratic equation along with this collected observations?

• 5.3, or 5.2? You've used both in the question. – Simon Hayward Oct 30 '12 at 14:19
• Oh! Sorry its $5.2 km$ – Saharsh Oct 30 '12 at 14:20
• Also, do you really need to form a quadratic to solve this? Have you been instructed to? Aren't you actually solving two simultaneous equations? – Simon Hayward Oct 30 '12 at 14:22
• Yep! It is compulsory to have a quadratic equation. And we have to solve the derived quadratic equation by factorisation or Discriminant formula – Saharsh Oct 30 '12 at 14:23
• Misread the first sentence. Have posted an appropriate answer now. – Simon Hayward Oct 30 '12 at 14:31

Ok, so we have $2 \text{(hours)} = \frac{5.2}{5-x}+\frac{5.2}{5+x}$, which rearranges simply enough into:- $2(5+x)(5-x)= 5.2(5+x)+5.2(5-x)$ from which you can gather terms to form a neat quadratic and solve for x.
• Can you explain, How $2 \text{(hours)} = \frac{5.2}{5-x}+\frac{5.2}{5+x}$, Rest are understood by me. – Saharsh Oct 30 '12 at 14:34
• @Simon I'd read this as $t_{downstream} = \frac{5.2}{5+x}$, $t_{upstream} = t_{downstream} + 1 = \frac{5.2}{5-x}$ which yields $\frac{5.2}{5-x} = \frac{5.2}{5+x} +1$ – roman Oct 30 '12 at 14:47