Quadratic is a type of problem which deals with a variable multiplied by itself- an operation also known as squaring.

Solution:

Let’s consider the speed of the fast train as x km/hr

Then, the speed of the slow train will be $=\left( x-10 \right)$ km/hr

Using, speed = distance/ time

Time taken by the fast train to cover $200$ km $=200/x$ hr

And, time taken by the slow train to cover $200$ km $=200/\left( x-10 \right)$ hr

Given, that the difference in the times is $1$ hour.

This can be expressed as below:

$\frac{200}{x}=\frac{200}{\left( x-10 \right)}=1$

$\frac{\left( 200\left( x-10 \right)-200x \right)}{x\left( x-10 \right)}=1$

$\frac{200x-2000-200x}{{{x}^{2}}-10x}=1$

${{x}^{2}}-10x=-2000$

${{x}^{2}}-10x+2000=0$

${{x}^{2}}-50x+40x+2000=0$ [by factorisation method]

$x\left( x-50 \right)+40\left( x-50 \right)=0$

$\left( x-50 \right)\left( x+40 \right)=0$

$x=50orx=-40$

As, the speed of train can never be negative we neglect $x=-40$

Thus, speed of the fast train is $50$ km/hr

And the speed of slow train $\left( 50-10 \right)=40$ km/hr