Question

A train travel a distance of $480km$ at a uniform speed. If the speed had been $8km/hr$ less, then it would have taken $3$ hours more to cover the same distance. Formulate the quadratic equation in terms of the speed of the train.

Answer 1

Let take speed of the train $=xkm/h$
And time taken by train in normal speed $=y$ hours
A train travels a distance of $480km$ at a uniform speed.
Distance $=$ speed $\times$ time
So we get
$x\times y=480\dots \dots \left(i\right)$
If the speed had been $8km/h$ less, then it would have taken $3$ hours more to cover the same distance
New speed $=x–8km/h$
Time taken $=y+3$ hours
Use same formula again
Distance $=$ speed $\times$ time
or, $(x-8)\times (y+3)=480$
or, $xy+3x-8y–24=480$
putting $xy=480$ from $eq.\left(i\right),$ we get
or, $480+3x–8y–24=480$
or, $3x–8y–24=0$
or, $-8y=24–3x$
or, $y=-3+\frac{3x}{8}$ [Divide both sides by $–8$]
on putting this value in equation $\left(i\right)$, we get
$x(-3+\frac{3x}{8x})=480$

or, $-3x+\frac{3x^2}{8}=480$

or, $-24x+3x^2=3840$ [Multiply both sides by $8$]

or, $x^2–8x=1280$ [Divide both sides by $3$]

or, $x^2–8x-1280=0$

It is in the form of $a{x}^2+bx+c=0$

where $a=1,b=-8,c=-1280$

Hence, the speed of the train in quadratic equation is $x^2–8x-1280=0$