Quadratic Equations

In elementary algebra, any equation of the form ax^2 + bx + c = 0 is termed as quadratic equation. In the equation, x represents an unknown value while a, b, and c represent known values. The value of a must not be 0.

In this post, you will learn how to solve quadratic equations and their usage in the real-world.

Solving Quadratic Equations

 

There are at least two popular ways of solving a quadratic equation but we will look at only one. The meaning of solving a quadratic equation is to find the value of x, the unknown value. In a quadratic equation there are two possible values of x and which one is correct depends on the real-world usage of the obtained value.

The general formula for finding x is:

    x1 = (-b + sqrt(b^2 - 4 * a * c)) / 2 * a
    x2 = (-b -  sqrt(b^2 - 4 * a * c)) / 2 * a

That's it! All it takes to solve a quadratic equation is plugging a, b, and c in the above formula.

Deriving The Formula

In this section, we will see how the formula presented above can be derived. Knowing this aspect will help you stand away from the crowd.

The equation is, ax^2 + bx + c = 0, where a is not equal to 0. Our goal is to isolate x, so move c to the right.

    ax^2 + bx = -c

Now, divide the whole equation by a.

    x^2 + bx/a = -c/a

Multiply bx/a with 2/2 (1). Multiplying by 1 doesn't change the value, so the equation is still valid.

    x^2 + 2b/2a * x = -c/a

You might be thinking why we are multiplying by 2/2. The reason is that we want the equation to come to this form: x^2 + 2xy + y^2 = (x + y)^2. Till now, we have x and are searching for y. Look carefully, y = b / 2a. We are only missing the y^2 term. Therefore, now we add y^2 (b / 2a)  to each side of the equation.

    x^2 + 2 * x * b/2a + (b / 2a)^2 = -c / a + (b / 2a)^2

Recall that we are trying to make the above equation fit this formula: x^2 + 2xy + y^2 = (x + y)^2. Now we can do this.

    (x + b/2a)^2 = b^2/4a^2 - c/a

Next, do the usual algebra to the terms on the right-hand side and you will get what is given below.

    (x + b/2a)^2 = (b^2 - 4ac) / 4a^2

Take the square root of both sides.

    x + b / 2a = sqrt(b^2 - 4ac) / 2a

If you are a mathematician, you might know that when you find the square root of d^2, you get |d| not d. Let me explain. What is the square root of 4? 2 and ... -2. You see, 2 * 2 = 4 and -2 * -2 = 4 as well. So, the square root of 4 is actually +/- 2 (either 2 or -2). Correcting the above equation, you will get this.

    x + b / 2a = +/- sqrt(b^2 - 4ac) / 2a

Recall that our objective in doing all this was to isolate x, and now we can do that.

    x = +/- sqrt(b^2 - 4ac) / 2a - b / 2a
>  x = +/- (sqrt0(b^2 - 4ac) - b) / 2a

That's it! You are done.

Real-World Usage of Quadratic Equations

The real-world usage of quadratic equations are varied and I am going to cover only one of them. Please note that whatever I say below is not 100 % accurate; the real-world calculations involved are very complicated and I have presented things in perhaps, the simplest way possible.

Without further lecture, let's jump straight in. Quadratic equations are used for calculating trajectories. A trajectory is the path taken by an object when it is thrown. Calculating trajectories allows one to determine accurately where a thrown object will fall or how an object should be thrown so that it hits the target. This finds immediate use in the military.

Because I don't have a missile handy, I will try to explain with the example of a ball; a golf ball to make it interesting :). Suppose you are on the Moon playing a game of golf. You put the ball on the stand which is 1 m high. You hit the ball straight up and the ball gained a velocity of 20 m/s. What time would the ball take to come down to the Moon's surface?

The initial height from the ground is 1 m.
The distance it covered in t time would be 20t m.
Gravity is pulling it down with a force of 2t^2 (force is represented as a square)

You can get the height by adding all these together. How? Simple, you started at 1 m, traveled 20t m and gravity added -2t^2 m to your ball's journey.

    -2t^2 + 20t + 1 = 0 (0 because that would be the distance from the ground when it would fall on the ground)

The above equation looks familiar. If it doesn't, to you, you need to sleep; staying awake for hours is not good.

Solve this and you will get: t = -0.04 and t = 10.04. Probably, the ball took 10.04 seconds to come back. Great, isn't it? No... then you will have to take one of those engineering courses.

If you have anything to say, let me know in the comments. Remember, knowledge increases with sharing, so go ahead and share this post with your friends.

See also:

Python Program to Find the Solutions of Quadratic Equations
Quadratic Equations On Wikipedia
Graphing Quadratic Equations
Complex Numbers On Wikipedia
Square Root of a Negative Number