• manogyana25
That equation will be quadratic because it will have x2. Solve that equation for x using the quadratic formula. You will get two solutions- one will be negative. The other will be the speed of the train.
manogyana25
Missing template due to originally being posted in non-homework forum
A train travels a distance of 480km at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. Find the speed of the train.
Represent this situation in the form of quadratic equation.

So I have been trying this but couldn't find it.
I have taken the speed as x.
And time, t
Distance =speed *time
Then
480=xt
480=(x-8)(t+3)
I couldn't go any further. And I'm not sure if this is right or not.
I just started learning quadratic equations. And I don't have any depth in the subject. Help me solve it in an easy way. Thank you!

manogyana25 said:
I have taken the speed as x.
Generally, avoid using x for speed as it is more regularly used for distances.

Apart from that (and the fact your quantities are missing units), you are doing well. You have two equations and two unknowns. Solve for t in one of them (I suggest the first) and insert the result into the other.

I have no problem with using "x" for speed! And in problems like this distance is more often "s" or "d".
(Though, personally, I would have used "v".)

manogyana25 said:
A train travels a distance of 480km at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. Find the speed of the train.
Represent this situation in the form of quadratic equation.

So I have been trying this but couldn't find it.
I have taken the speed as x.
And time, t
It would be better to say that t is the time required to travel 480 km at the original speed. Just "t is time" is ambiguous. And it would be better to say "x is the speed in km per hour and t is in hours" but those are a bit picky!

Distance =speed *time
Then
480=xt
480=(x-8)(t+3)
I couldn't go any further. And I'm not sure if this is right or not.
I just started learning quadratic equations. And I don't have any depth in the subject. Help me solve it in an easy way. Thank you!
From 480= xt,, t= 480/x. Replace t in the second equation with that to get a single equation in x.

## 1. What is a quadratic equation?

A quadratic equation is an equation of the form ax2 + bx + c = 0, where a, b, and c are constants and x is the variable. It represents a polynomial with a degree of two, and its graph is a parabola.

## 2. How do I solve a quadratic equation?

To solve a quadratic equation, you can use the quadratic formula: x = (-b ± √(b2 - 4ac)) / 2a. You can also use factoring or completing the square methods. It's important to remember to check your solutions by plugging them back into the original equation.

## 3. What are the different types of solutions for a quadratic equation?

There are three types of solutions for a quadratic equation: real, imaginary, and complex. Real solutions are when both roots are real numbers, imaginary solutions are when the discriminant (b2 - 4ac) is negative and the roots are imaginary numbers, and complex solutions are when the discriminant is zero and the roots are complex numbers.

## 4. What is the discriminant of a quadratic equation and how is it used?

The discriminant of a quadratic equation is the expression b2 - 4ac. It is used to determine the nature of the solutions of the equation. If the discriminant is positive, the equation has two real solutions. If it is negative, the equation has two imaginary solutions. And if it is zero, the equation has one real solution.

## 5. How can quadratic equations be applied in real life?

Quadratic equations have many real-life applications, such as in physics, engineering, and finance. For example, they can be used to model the trajectory of a projectile, the shape of a bridge arch, or the profit of a business. They can also be used to find the maximum or minimum value of a function, which is useful in optimization problems.

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