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Hint- Here, we will be using distance-speed-time formula.

Let the actual uniform speed of the train be x km per hour and the actual time taken by the train to cover 480 km be hours.

As we know that ${\text{Speed}} = \dfrac{{{\text{Distance}}}}{{{\text{Time}}}}$

First Case- When the train is covering 480 km at a uniform speed of \[\] km per hour and time taken by the train is $t$ hours then,

$x = \dfrac{{480}}{t} \Rightarrow t = \dfrac{{480}}{x}{\text{ }} \to {\text{(1)}}$

Second Case- When the train is covering 480 km at a speed of $\left( {x - 8} \right)$ km per hour and time taken by the train is $\left( {t + 3} \right)$ hours then,

$\left( {x - 8} \right) = \dfrac{{480}}{{\left( {t + 3} \right)}} \Rightarrow \left( {x - 8} \right)\left( {t + 3} \right) = 480\,{\text{ }} \to {\text{(2)}}$

In order to get the final quadratic equation also in terms of speed of train so we will eliminate variable $t$ in the above equation.

Using equation (1) in equation (2), we get

$\

\Rightarrow \left( {x - 8} \right)\left( {\dfrac{{480}}{x} + 3} \right) = 480\, \Rightarrow \left( {x - 8} \right)\left( {\dfrac{{480 + 3x}}{x}} \right) = 480\, \Rightarrow \left( {x - 8} \right)\left( {480 + 3x} \right) = 480x\, \\

\Rightarrow 480x + 3{x^2} - 3840 - 24x = 480x \Rightarrow 3{x^2} - 24x - 3840 = 0 \Rightarrow 3\left( {{x^2} - 8x - 1280} \right) = 0 \\

\Rightarrow {x^2} - 8x - 1280 = 0 \\

$

The above quadratic equation is in terms of speed of the train i.e., variable $x$.

Note- In this particular problem, we have two variables (speed and time taken by the train). According to the problem statement, we are given with two conditions and these conditions can be formulated into two equations in two variables but it is asked in the problem to represent the final quadratic equation in terms of one variable (speed) only so, for that we will express the other variable (time) in terms of the variable (speed) finally needed in the answer.

Let the actual uniform speed of the train be x km per hour and the actual time taken by the train to cover 480 km be hours.

As we know that ${\text{Speed}} = \dfrac{{{\text{Distance}}}}{{{\text{Time}}}}$

First Case- When the train is covering 480 km at a uniform speed of \[\] km per hour and time taken by the train is $t$ hours then,

$x = \dfrac{{480}}{t} \Rightarrow t = \dfrac{{480}}{x}{\text{ }} \to {\text{(1)}}$

Second Case- When the train is covering 480 km at a speed of $\left( {x - 8} \right)$ km per hour and time taken by the train is $\left( {t + 3} \right)$ hours then,

$\left( {x - 8} \right) = \dfrac{{480}}{{\left( {t + 3} \right)}} \Rightarrow \left( {x - 8} \right)\left( {t + 3} \right) = 480\,{\text{ }} \to {\text{(2)}}$

In order to get the final quadratic equation also in terms of speed of train so we will eliminate variable $t$ in the above equation.

Using equation (1) in equation (2), we get

$\

\Rightarrow \left( {x - 8} \right)\left( {\dfrac{{480}}{x} + 3} \right) = 480\, \Rightarrow \left( {x - 8} \right)\left( {\dfrac{{480 + 3x}}{x}} \right) = 480\, \Rightarrow \left( {x - 8} \right)\left( {480 + 3x} \right) = 480x\, \\

\Rightarrow 480x + 3{x^2} - 3840 - 24x = 480x \Rightarrow 3{x^2} - 24x - 3840 = 0 \Rightarrow 3\left( {{x^2} - 8x - 1280} \right) = 0 \\

\Rightarrow {x^2} - 8x - 1280 = 0 \\

$

The above quadratic equation is in terms of speed of the train i.e., variable $x$.

Note- In this particular problem, we have two variables (speed and time taken by the train). According to the problem statement, we are given with two conditions and these conditions can be formulated into two equations in two variables but it is asked in the problem to represent the final quadratic equation in terms of one variable (speed) only so, for that we will express the other variable (time) in terms of the variable (speed) finally needed in the answer.

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