Quadratic what?!

King of the Hill rocks!! Yep, I just finished watching the series. It's kinda funny but yet, it's amazing how the creators of Beavis and Butthead could come up with a story that's something funny and crazy (considering the fact that B&B always ends up doing more harm than good in the end) and then ice things up with a mellow tone that displays life's practicality - that is, regardless of the stupidities and craziness that they do, they always end up settling for what is really the right thing. It's like a skateboard punk guy that acts rugged and all but deep down inside, he still finds time to help his mom shop at the mall or feed his grandma with some cereal and milk. It's really a nice cartoon - in fact, one of the nicest yet funny cartoon out there - and I'd recommend that you'd watch it.

Anyway, I just finished working on the first 20 questions at the end of chapter 1.6 and I came across this problem that required me to solve using a quadratic equation. I was kinda excited to post this in here so, here it goes..

Basically, there are two common types of equation that you'll encounter out there; one is a linear equation and the other one is a quadratic equation. Linear equations are those equations that contains a constant and a first-order term. In other words, no variable can be raised to the power other than 1. An example would be 3x = y or y= ax + b (where a and b are constants). Solving for x would be easy as you only need to separate x from other variables and coefficients by following basic algebra theories. Ever wonder why they call it linear equation? That is beacuase of the fact that a linear equation is also the equation of a straight line. It means that if you continuously solve for both x and y, you end up with a pair of numbers that when plotted on a cartesian plane, results in points having the same slope as the previous plotted points - meaning a straight line. Anyway, the other equation is the quadratic equation. Unlike the linear, this equation has a second-order term and a constant. This means that it contains a variable that is raised to the power of 2. An example would be x2 - 6x + 8 = 0. Solving for x now involves a lot more than solving for x in a linear equation. A gentelman by the name of Gauss once proved that an equation having a degree of 2 or x has 2 or x number of solution depending on the number of the degree. Hence, in the equation x2 - 6x + 8 = 0, we both have numbers 4 and 2 that will satisfy an equality (try it!). Solving for x in a quadratic type of equation can either be: 1) Completing the square of the equation by adding b/2 on both sides of the equation. 2) Using the quadratic formula. 3) Factoring the equation (if it is factorable). I find methods 1 and 2 very convenient especially for complex quadratics. So why the hell did they name this type of equation a quadratic equation anyway? That, I do not konw :)

So, you might be asking "Where the hell am I gonna use a quadratic equation anyway?!" well, I did ask that myself but that was before I read chapter 1.6 :) Here's a simple example that demonstrates the use of the quadratic equation:

Find two numbers whose sum is 70 and product 1224.

Without the knowledge of Algebra, we are doomed to guess random numbers and hope that their sum is 70 and their product is 1224. Luckily, we know enough than just doing random guess. So:

Let:
First Number = x
Second Number = y

The sum of the First Number and the Second Number is 70.
x + y = 70

The product of the First Number and the Second Number is 1224.
x . y = 1224

We have two unknowns in the form of x and y so we have to eliminate one unknown in order for us to solve the equation and we can do that by expressing one unknown in terms of the other.

Hence:
x + y = 70
x = 70 - y

Expressing x in terms of y, we have:
x . y = 1224
(70 - y) . y = 1224
y2 -70y + 1224 = 0

Now, we have arrived at a quadratic equation by simplifying the equation. So, we can use either of the methods mentioned above in solving this equation. In this case, let's use the quadratic formula in solving the problem.

The quadratic formula has a general form of:

(-b +- sqrt(b2 - 4ac)) / 2a

(sorry but this blog editor doesn't support square root and superscripts)

Substituting the values to the formula, we have:
(-(-70) +- sqrt( (-70)2 - 4(1)(1224))) / 2(1)

Since we have two solutions for the equation, we then have:

Solution 1:
(70 + sqrt(4) )/2

Solution 2:
(70 - sqrt(4) )/2

Which will give us:

Solution 1:
36

Solution 2:
34

So does this satisfy the original problem? Let's see:

The sum of the First Number and the Second Number is 70.
x + y = 70
36 + 34 = 70
70 = 70

The product of the First Number and the Second Number is 1224.
x . y = 1224
36 . 34 = 1224
1024 = 1224

We have an equality!! And so without having to guess a lot of random numbers, we have arrived at the answers to the equation without even having to break a sweat! This is an easy one. There are a lot of problems out there in the world that requires much complex computation than this one. But of course, knowing the fundamentals really simplify matters a lot!! I'm planning to do a Java applet that does a simulation which incorporates solving equations (like those work rate problem that we had on my last blog) - but that is after I learn how to do applets. I'm planning to touch it next week though.

I'll be continuing to solve 10-20 more numbers on the book today before I sleep. So I guess this is it for now. I'm gonna hit the books again and hopefully, I don't get papercuts this time ;)