Question

Note: This is a tutorial, NOT a question.
Solving the quadratic equation.

Answer 1

There are 3 main ways to solve the quadratic equation. Each presents its advantages. We will first be discussing factoring, then completing the square, and finally solving by formula.

Answer 2

A quadratic equation is an equation in the form
\(\huge y=f(x)=ax^2+bx+c\)

Answer 3

Now, what are \(y\) and \(f(x)\) you ask? Simple. When we put in a value for x, we get out a value y, sometimes also called f(x).

And what are a and b and c? a and b and c are constants. You may be wondering, if they are constants, why must we show them as variables? The answer is that although a, b, and c are constants when we are given an equation, when we describe a quadratic equation, it is easier to give these three constants a general form - when we say y=ax^2+bx+c we are talking about the quadratic equation in general. When we talk about y=3x^2+5x-9 we are talking about that specific quadratic equation.

Answer 4

So, why does y or f(x) always seem to equal 0? Well, the truth is, it doesn't. In fact, we set it to zero, because we usually try to find the "zeros" of the equation, which just means we try to find out for what values of x is y zero. On the graph, this is the same thing as when the graph intersects the x-axis.

Answer 5

Then, how does one solve a quadratic equation by factoring? Go to the link below for the answer :)

Answer 6

http://openstudy.com/users/lgbasallote#/updates/4f9bd87ee4b000ae9ed11143
Courtesy of @lgbasallote . Please give *him* a medal if you find his guide helpful.

Answer 7

Now that we have learned how to solve the quadratic equation by factoring, we shall learn how to complete the square. Many people at this point might wonder why we even learn another method. It's simple. We can't actually solve all quadratic equations by factoring. If you doubt me, I'll give you an examples.
x^2-2x-2=0
3x^2+10x+2=0
In fact, in any quadratic equation, if \(\huge \sqrt{b^2-4ac}\) is not rational, you will have difficulty factoring the equation.

Answer 8

So what is this "completing the square"? It's actually very simple, once you get the idea. The very basic concept of completing the square is that we want to write
\(ax^2+bx+c=0\) in the form \((x+u)^2+v=0\); we don't exactly know what u and v are, but I will soon show you how to find them, as well as make evident that they definitely exist.

Answer 9

To get a basic idea of how completing the square works, let's square (x+u)
\((x+u)^2=x^2+2xu+u^2\)
This seems useless, but now we know that if we have an equation in the form
\(x^2+2ux+u^2\) we can take the square root (or factor, if you would like) it into
\((x+u)^2\)

Answer 10

So, why don't we solve some equations in this form first, just to get a feel? I'll provide a few examples, and solve each one.
1. \(x^2+6x+9=0\)
2. \(x^2-4x+4=0\)
3. \(2x^2+4x+2=0\)
4. \(x^2+x+\frac{1}{4}=0\)
5. \(3x^2+9x+\frac{27}{4}=0\)

Answer 11

So, let's look at the first problem.
We can rewrite it into
\(x^2+2(3x)+3^2=0\)
Huh. That looks familiar.
Why don't we just take the square root and get
\(\huge (x+3)^2\)? There's two ways to check if I'm right. The first is to use FOIL to multiply it out. The second is to use factoring, more specifically, factoring with the formula (a+b)^2=a^2+2ab+b^2. In fact, we're simply just doing the reverse of that.

Answer 12

Now, let's look at the second problem.
\(x^2-4x+4\)
Let's just simply rewrite it into
\(x^2+2(-2x)+(-2)^2\)
Same stuff, only a negative instead of a positive.
This is actually quite important. Remember (a-b)^2=a^2-2ab+b^2? Well, this is the reverse of that.

Answer 13

How about the third problem?
Well, this one is a little tricky if you don't know what you need to do for the first step.
We divide both sides by 2. Remember, 0 divided by any number (but 0 of course) is still 0, so if we divide both sides by 2, it's completely valid.
\(\huge \frac{2x^2+4x+2}{2}=\frac{0}{2}\)
=> \(x^2+2x+1=0\)
Hopefully you recognize this as
\((x+1)^2=0\)
Why did we divide by 2? The answer is that we divided by 2 to make the coefficient of x^2 (the number in front of x^2, or, 2), 1. We just got rid of that number in front of the x^2 to make completing the square easier.

Answer 14

Number 4
\(x^2+x+\frac{1}{4}=0\)
This one looks hard, but why don't we try doing what we've done in the past?
\(x^2+2*(\frac{1}{2}x)+\sqrt{\frac{1}{4}}^2=0\)
That makes it significantly easier.
\(x^2+2(\frac{1}{2}x)+\frac{1}{2^2}\)
In \((x+b)^2\), just set b to 1/2
-->\((x+\frac{1}{2})^2=0\)
And there we have it.

Answer 15

Finally, number 5. In this problem, we will apply what we did in number 3, then number 4.
So, the problem is \(\huge 3x^2+9x+\frac{27}{4}=0.\)
First, let's just divide by what's in front of x^2, to get it out of the way. In this case, that would be 3. The result is
\(x^2+3x+\frac{9}{4}=0.\)
Now, rewrite it.
\(x^2+2*(\frac{3}{2}x)+\sqrt{\frac{9}{4}}\)
Or,
\(\large x^2+2*(\frac{3}{2}x)+\sqrt{\frac{3^2}{2^2}}\)
So, for (x+b)^2 we have b=3/2.
Then, we simplify to
\(\huge (x+\frac{3}{2})^2\)
And we (I) am done with those examples.

Answer 16

Now the big question is "Can one actually do this to all equations?"
The answer is sort of.

The equation must be in the form that we can just take the square root of it like that to solve it. However, many equations, including the ones that I mentioned were unsolvable by factoring (and additionally most that can be solved by factoring), aren't capable of just being taken the square root of.
However, we can do something that makes it so we can take the square root of them. This is known as "completing the square".

We shall now work through those "unsolvable by factoring" problems by completing the square.

Answer 17

1. \(\large x^2-2x-2=0\)
2. \(\large 3x^2+10x+2=0\)

Answer 18

Number one:
Now, I said that we can make the equation in the form we want, but it seems like we can't touch the x^2 term, or the -2x term. So let's not. How about we \(\large ignore\) that -2, and try to add on that term to complete the square?

\(x^2-2x+...=0\) -> resembles \(x^2-2x+1\), which we know to be \((x-1)^2\)
So, let's just add one to both sides.
\(x^2-2x+1+...=0+1\) But now, we need to pay attention to that -2, because when we put it in, it gets us nowhere. Why don't we just move it to the other side?
\(x^2-2x+1+(-2+2)=0+1+2\)
\(x^2-2x+1-0=3\)
\(x^2-2x+1=3\)
Now, let's rewrite the \(x^2-2x+1\) to \((x-1)^2\)
\((x-1)^2=3\) -> take the square root of both sides. (Remember, there are two square roots. Positive and negative. if you don't understand, just make a comment, and I'll explain.)
\((x-1)=\pm \sqrt{3}\)
So, \(x=1\pm \sqrt{3}\)
We could write both solutions separately.
\(x=1+ \sqrt{3}\), and
\(x=1- \sqrt{3}\)

Answer 19

\(3x^2+10x+2=0\) Now, since I'm lazy and you're knowledgeable, I'll just show my work, and you'll understand it without an explanation.
\(\large x^2+\frac{10}{3}x+\frac{2}{3}=0\)
\(\large x^2+2*\frac{5}{3}x+\frac{2}{3}=0\) ->get rid of the 2/3 by moving it to the other side.
\(\large x^2+2*\frac{5}{3}x=-\frac{2}{3}\) ->complete the square. Make sure to add to both sides.
\(\large x^2+2*\frac{5}{3}x+\frac{25}{9}=-\frac{2}{3}+\frac{25}{9}\)
Simplify.
\(\large (x+\frac{5}{3})^2=\frac{19}{9}\)
Take the square root of both sides.
\(\large x+\frac{5}{3}=\pm \frac{\sqrt{19}}{3}\)
Isolate x.
\(\large x=-\frac{5}{3}\pm \frac{\sqrt{19}}{3}\)
And there's your answer. If we really wanted to, we could write both solutions of x out.
\(\large x=-\frac{5}{3}+ \frac{\sqrt{19}}{3}\)
\(\large x=-\frac{5}{3}- \frac{\sqrt{19}}{3}\)
And, that's it.

Answer 20

Now, I told you that we can solve equations that are not factorisable with the "completing the square" method. Well, we can actually solve all quadratic equations with completing the square method. We start with our general equation \(ax^2+bx+c=0\) and complete the square. What comes up at the end is a handy formula that can be used for any quadratic. You might not understand the work behind it, but try to look over it; someday you may come back and find it makes perfect sense.

Answer 21

We start with the equation.
\(\Huge ax^2+bx+c=0\)
\(\Huge x^2+\frac{b}{a}*x+\frac{c}{a}=0\)
(getting rid of what's in front of x.)
\(\Huge x^2+\frac{b}{a}*x=-\frac{c}{a}\)
(getting rid of that constant)
\(\Huge x^2+2*\frac{b}{2a}*x=-\frac{c}{a}\)
(rewriting the equation)
\(\Huge x^2+2*\frac{b}{2a}*x+(\frac{b}{2a})^2=-\frac{c}{a}+(\frac{b}{2a})^2\) (adding what we need)
\(\Huge (x+\frac{b}{2a})^2=(\frac{b}{2a})^2-\frac{c}{a}\)
(re-writing as a perfect square)
\(\Huge (x+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{c}{a}*\frac{4a}{4a}\)
(simplifying and rewriting as a common denominator.)
\(\Huge (x+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}\)
\(\Huge (x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\)
(rewriting into a nicer form)
\(\Huge x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}\)
(taking the square root)
\(\Huge x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}\)
(simplifying)
\(\Huge x=-\frac{b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}\)
(isolating the variable)
\(\Huge x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
(and we're done!)

Answer 22

Line 5 is cut off. It is missing \(\huge )^2\) at the very edge.

Answer 23

This part is not easy to explain, but I will try my best. A quadratic equation either has one repeated solution, two different solutions, or no real solutions (2 complex conjugate solutions, for those required to learn about imaginary and complex numbers) Now, there is a more graphical interpretation of the solutions, as well as another way to tell how many solutions the equation has by knowing \(\huge a,b,c\).
I will try to draw graphs to help explain.

Answer 24

Case 1: two different solutions.
Whenever we set y to 0, we are saying that it crosses the x axis. Why? Remember how in a system of equations, the intersection of the two equations was the solution? This is the same. These are the two equations.
y=ax^2+bx+c
y=0 (the equation for the x-axis.)
Whenever we set to 0, it becomes
0=ax^2+bx+c
The intersections with the x-axis (y=0) are the solutions. In this case, there are two.