Quadratic equations

Miscellaneous Topics
#1

Hi all, I’m trying to learn how to solve quadratic equations but I’ve ran into a problem.
On the wikipedia page: http://en.wikipedia.org/wiki/Quadratic_equation
It explains the step by step process of solving a quadratic equation using the “Complete the square” method. The fourth step states: “Write the left side as a square, and simplify the right side, if necessary.”
And this appears to change the expression “x² + 2x + 1” into “(x + 1)²” and I’m not exactly sure how that’s come about. Could somebody explain it to me?

Thanks!

Paul

#2

If its for a program, just use the quadratic equation that solves it all:

x1,x2 = (-b (+/-) sqrt(b^2-4ac)) /2a

you have to do it twice, one for the + and another for the -

#3

Basically,you multiply (x+1) by (x+1). So you go
First, (x+1)(x+1), which is xx, or x^2
Outside, (x+1)(x+1), which is x1, or simply x
Inside, (x+1)(x+1), which is 1x, or simply x
Last, (x+1)(x+1), which is 11, or 1
Then you put it together and it becomes x^2 + x + x + 1, which is then simplified to x^2 + 2x + 1. This process is called FOIL.

#4

Thanks for the replies guys! So Geemili, is what I described just FOIL in reverse?

#5

Oh, I didn’t read your post closely enough. I don’t know the process of getting from “x² + 2x + 1”, how I remember solving it in math class is by plugging in random numbers into (ax + b)(cx + d), until it would form the quadratic equation through FOIL. I’m sure there’s a better way, I just don’t know it. :stuck_out_tongue:

#6

Yeah, I keep re-reading this instructions thinking I’m missing a subtle wording, but that step completely throws me. The other steps are fairly trivial.

#7

Holy crap you don’t know Algebra (yet hopefully?)…how old are you Porlus?!

#8

Old enough for it to be embarrassing. ): I’m alright with other areas of algebra. This has just stumped me.

#9

I’ll second what Agro said about the quadratic formula. It’s really easy, you just plug in the numbers and out comes the answer.
(-b±√(b²-4ac)) / (2a)
As far as I remember you use this with a trinomial or something like that… I haven’t used it in a while. =P

#10

Thanks for the reply. :slight_smile: Yeah, all I remember from secondary school with regards to quadratic equations was that formula. I’m just trying to look at the other methods of solving it to help me get a better understanding. Completing the square is the method that I recall the teacher saying takes too long to work out in an exam. Hurray for the British education system. D: So now I’m trying to learn it on my own and struggling a bit.

#11

The “plug in numbers” method is bullshit, but all the schools teach it. I don’t think that’s his question though…
“x² + 2x + 1”
To be able to simplify, your equation needs to be in the form a^2 + 2ab + b^2 which it already is. Now “a^2 + 2ab + b^2” simplifies into “(a+b)^2”, so looking at your equation, the root of x² is x and the root of 1 is 1. Thus you have (x+1)^2.

#12

I dot know what your talking about. The quadradic formula is great for getting values of x that aren’t whole for polynomials with degrees of two. It doesn’t help in factoring at all, but if you already have x, why factor?

#13

He’s asking someone to explain why “x² + 2x + 1” can be factored into “(x + 1)²” and how. He never asked anything about finding what x was, or how to use it in a program.

#14

Solving Quadratic Equations (by brain):

You have to make one side equal zero. Because if one side is zero, one (or both) of the factors has to be zero. Then you find the 2 (or sometimes 1) value(s) of x, and you take the one you want.

1. If the equation is in the form (ax + b)^2 = 0

This means that ax + b = 0
-> ax = -b
-> x = (-b)/a

2. If the equation is in the form (ax + b)(cx + d) = 0

This means that either ax + b = 0 or cx + d = 0
First, assume ax1 + b = 0 (x1 and x2 are the possible values of x)
-> x1 = (-b)/a
and then cx2 + d = 0
-> x2 = (-d)/c

So now you have x1 and x2. But you only want one. This is where it comes down to common sense. (eg: can’t have negative amounts of people etc.) Look back to the situation.

3. If the equation is in the form ax^2 + bx + c = 0

You need to get this into factorised form.

Find two numbers (d and e) that match the following equations:
de = ac (and therefore: d/a = e/c, which is important later)
d+e = b

Now you can rearrange the formula to this:
ax^2 + dx + ex + c = 0

Factorise the first two terms, and the last two terms seperately:
ax^2 + dx = ax(x + d/a)
ex + c = e(x + c/e)

And because d/a = c/e (as shown above), we’ll make it easier by having f = d/a = c/e.
The new equations:
ax(x + f)
e(x + f)

And now to add them together again:
ax(x + f) + e(x + f) = 0
(ax + e)(x + f) = 0

Now continue with method #2 above.

4. For other equations

Rearrange into ax^2 + bx + c = 0, then use method #3.

Note: For programming purposes, use the quadratic formula.

#15

It doesn’t take that long, and it’s the way that the formula you memorise was derived.

If you’re needing to solve quadratic equations for a game, be aware that there are numerical issues with one half of -b +/- sqrt(b^2 - 4ac) when 4ac is small. The way to handle this is to have a case split on the sign of b and then to use the fact that the product of the roots is c/a.

#16

Completing the square is way easier if the variable a in the quadratic equation does not have coefficients, e.g. x2 + 3x + 2. Then it’s by far way simpler than having to write out the whole quadratic formula and solve it. But if you have a coefficient for the variable a, it takes way too long.

#17

Ah, I understand now. I guess just the way you worded it made me think differently.

#18

Not necessarily true. The multiples of the coefficient of “a” you can plugin on the left of a matrix - and the multiples of “c” you can plug in on the right. For example: 5x2 + 12x + 4

5|2

1|2

Cross multiply and add 52 + 21 = 12. Which means this is the solution, because it equals the coefficient of x (middle term) - to convert into 2 binomials just do (5x + 2)(x + 2). Obviously, you still have to plug in some numbers, but this way is a lot faster.