Jump to content
• Sign Up

# Roots

## Recommended Posts

For finding the roots,

like

root(x^2),

What's the value?

Is that if

x > 1, ans=x

x<1, ans=-x

x=0, ans=0?

But when we do quandratic equations,

(x^2)=4

x =2 or x=-2?

Should we rearrange the steps as

x^2=4

x=2 or -x=2?

#### Share this post

##### Share on other sites

It doesn't really matter, since $x=-2$ and $-x=2$ are the same equation. To show that this is true, just multiply both sides of the first equation by $-1$.

#### Share this post

##### Share on other sites

However, there is the surds are defined as positive.

Is that only for calculating roots?

#### Share this post

##### Share on other sites

Reading your post I would assume that you mean the following

First Part

$f(x) = \sqrt {x^2}$

Hence $f(x) = x}$

Therefore x has the domain of $(-\infty , \infty )$

Second Part

The problem is to find the real zeros of a linear function f(x)=0

$f(x) = x^2 - 4$

$(x -2)(x + 2) = 0$

$x = 2, x = -2$

#### Share this post

##### Share on other sites

if f has the domain you claim, Gauss, then it isn't the identity function, since roots are always taken to be positive.

#### Share this post

##### Share on other sites
$f(x) = \sqrt {x^2}$

Hence $f(x) = x}$

Could be wrong here' date=' but I think that [math']f(x) = |x|[/math].

#### Share this post

##### Share on other sites
Could be wrong here, but I think that $f(x) = |x|[/math']. If I remember, there were two sides to this. I was doing a problem and trying to be a little creative. It asked you to write [math]|x|+|x+3|$ but without the absolute values. I knew they wanted me to restrict the domains and such to get three different equations that would all produce the desired result. But aplying the example problems right from the chapter to the homework questions is so boring. So I decided to not read the chapter and try and derive some of the solutions on my own. For this one, I rewrote the equation as $\sqrt{x^2}+\sqrt{x^2+3}$. When graphed, made into a table or whatever you want, they always yeild the same solution. So I asked the teacher why that was not an equaly valid answer. She said that I found the technical definition of absolute values, but that writing the equation that way causes problems because people would assume that you should reduce the power and the root to leave you with just $x+x+3$(or not, but you get the point) which doesn't work. I still don't know exactly what's going on with this stuff. Does anyone know whether mine should be equaly valid or should be discarded?

#### Share this post

##### Share on other sites

The function f(x) = x is an identity function.

When expressions have more than one operation, we follow rules for the order of operations. PEMDAS

1. Parentheses

2. Exponents

3. Multiplication and Division

4. Addition and Subtraction

$f(x) = \sqrt{x^2}$

$f(x) = x^{\frac {2}{2}$

$f(x) = x$

Case when x = -3

$f(x) = \sqrt{-3^2}$

$f(x) = -3^{\frac {2}{2}$

$f(x) = -3$

Case when x = 3

$f(x) = \sqrt{3^2}$

$f(x) = 3^{\frac {2}{2}$

$f(x) = 3$

Another way

Case when x = -3

$f(x) = \sqrt{-3^2}$

$f(x) = \sqrt{9}$

$f(x) = 3$

Case when x = 3

$f(x) = \sqrt{3^2}$

$f(x) = \sqrt{9}$

$f(x) = 3$

#### Share this post

##### Share on other sites
Case when x = -3

$f(x) = \sqrt{-3^2}$

$f(x) = \sqrt{9}$

$f(x) = 3$

That's not the same as saying that $\sqrt{x^2}$ is equal to $x$

#### Share this post

##### Share on other sites

So, Gauss, f(x)=x and then you show by example that f(-3) is not equal to -3.

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account

## Sign in

Already have an account? Sign in here.

Sign In Now

×

• #### Activity

• Leaderboard
×
• Create New...

## Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.