# [SOLVED]Term under root

#### wishmaster

##### Active member
It was given this term:

$$\displaystyle \sqrt{1-(x^2+1)^2}$$

My friend got the solution $$\displaystyle -x^2$$,but i think he is not right a about that,cause i believe you should first solve the quadratic binom,which is $$\displaystyle x^4+2x+1$$,so my solution is:
$$\displaystyle \sqrt{1-x^4-2x-1}$$ or $$\displaystyle \sqrt{-x^4-2x}$$

Im wondering who is right now,or how to solve this. Thank you!

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#### MarkFL

Staff member
It was given this term:

$$\displaystyle \sqrt{1-(x^2-1)^2}$$
What are you supposed to do? Find the domain? Simplify?

#### wishmaster

##### Active member
What are you supposed to do? Find the domain? Simplify?
Simplify,but i have made corrections,so take a look.

#### wishmaster

##### Active member
Simplify,but i have made corrections,so take a look.
Actualy,original question was:

$$\displaystyle f(x)=x^2+1$$ and $$\displaystyle g(x)=\sqrt{1-x^2}$$

Find the composition $$\displaystyle (g\circ f)(x)$$

#### MarkFL

Staff member
Actualy,original question was:

$$\displaystyle f(x)=x^2+1$$ and $$\displaystyle g(x)=\sqrt{1-x^2}$$

Find the composition $$\displaystyle (g\circ f)(x)$$
$$\displaystyle (g\circ f)(x)=g\left(f(x) \right)=\sqrt{1-\left(x^2+1 \right)^2}$$

Can you finish?

#### wishmaster

##### Active member
$$\displaystyle (g\circ f)(x)=g\left(f(x) \right)=\sqrt{1-\left(x^2+1 \right)^2}$$

Can you finish?
As i wrote:

$$\displaystyle \sqrt{1-(x^4+2x+1)}$$ which is $$\displaystyle \sqrt{1-x^4-2x-1}$$.equals to $$\displaystyle \sqrt{-x^4-2x}$$ or am i wrong?

#### MarkFL

Staff member
As i wrote:

$$\displaystyle \sqrt{1-(x^4+2x+1)}$$ which is $$\displaystyle \sqrt{1-x^4-2x-1}$$.equals to $$\displaystyle \sqrt{-x^4-2x}$$ or am i wrong?

$$\displaystyle \left(x^2+1 \right)^2$$

#### wishmaster

##### Active member

$$\displaystyle \left(x^2+1 \right)^2$$
$$\displaystyle x^4+2x^2+1$$ ??

ok,i have missed a power on $$\displaystyle 2x$$

but then comes $$\displaystyle \sqrt{-x^4-2x^2}$$ I think i cant simplify that.....or?

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#### Prove It

##### Well-known member
MHB Math Helper
$$\displaystyle x^4+2x^2+1$$ ??

ok,i have missed a power on $$\displaystyle 2x$$

but then comes $$\displaystyle \sqrt{-x^4-2x^2}$$ I think i cant simplify that.....or?
You can simplify it if you take out a factor of \displaystyle \begin{align*} x^2 \end{align*} and remember that \displaystyle \begin{align*} \sqrt{x^2} = |x| \end{align*}.

#### wishmaster

##### Active member
You can simplify it if you take out a factor of \displaystyle \begin{align*} x^2 \end{align*} and remember that \displaystyle \begin{align*} \sqrt{x^2} = |x| \end{align*}.
So then:

$$\displaystyle \sqrt{-x^2(x^2+2)}$$
or $$\displaystyle x\sqrt{x^2+2}$$ ??

#### Prove It

##### Well-known member
MHB Math Helper
So then:

$$\displaystyle \sqrt{-x^2(x^2+2)}$$
Yes

or $$\displaystyle x\sqrt{x^2+2}$$ ??
Definitely not! Did you not read my previous post? Also, where have the negatives gone inside your square root?

#### wishmaster

##### Active member
Can you show me how?

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#### Jameson

Staff member
Can u show me how?
Hi wishmaster!

I don't know who "u" is, but maybe "you" can find him for us .

Try it this way. We have $\displaystyle \sqrt{-x^2(x^2+2)}$. Let's distribute the negative first. This becomes $\displaystyle \sqrt{x^2(2-x^2)}$. What about now?

EDIT: Should be $\displaystyle \sqrt{x^2(-2-x^2)}$ as Pranav states below.

#### Pranav

##### Well-known member
Try it this way. We have $\displaystyle \sqrt{-x^2(x^2+2)}$. Let's distribute the negative first. This becomes $\displaystyle \sqrt{x^2(2-x^2)}$. What about now?
I guess you meant $\displaystyle \sqrt{x^2(-2-x^2)}$.

#### wishmaster

##### Active member
Hi wishmaster!

I don't know who "u" is, but maybe "you" can find him for us .

Try it this way. We have $\displaystyle \sqrt{-x^2(x^2+2)}$. Let's distribute the negative first. This becomes $\displaystyle \sqrt{x^2(2-x^2)}$. What about now?
I apologize for "u",i know the rules,but sometimes i forget....sorry-1

And now im stuck with this term......

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I guess you meant $\displaystyle \sqrt{x^2(-2-x^2)}$.
I apologize for "u",i know the rules,but sometimes i forget....sorry-1

And now im stuck with this term......
Consider that generally:
\begin{array}{lcl}
\sqrt{a \cdot b} &=& \sqrt a \cdot \sqrt b \\
\sqrt{a^2} &=& |a|
\end{array}
Can you apply those rules and bring $|x|$ properly outside the square root?

There is one more step that I'll get to after this one.

#### Jameson

Staff member
I guess you meant $\displaystyle \sqrt{x^2(-2-x^2)}$.
Doh! You're quite right. That'll teach me to do math just after waking up.

#### wishmaster

##### Active member
Consider that generally:
\begin{array}{lcl}
\sqrt{a \cdot b} &=& \sqrt a \cdot \sqrt b \\
\sqrt{a^2} &=& |a|
\end{array}
Can you apply those rules and bring $|x|$ properly outside the square root?

There is one more step that I'll get to after this one.
$$\displaystyle x\sqrt{-2-x^2}$$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
$$\displaystyle x\sqrt{-2-x^2}$$
That should be $|x| \sqrt{-2-x^2}$.
Not sure why you would leave out the absolute value symbols.
Oh. Perhaps you do not know what those are?

Let me clarify.
The absolute value of a number x is the magnitude of that number.
One might also say: the number without its sign.
It is written as $|x|$ and can be defined as:
$$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$$

Moving on.
Are you aware that a square like $x^2$ is always either positive or zero, but cannot be negative?
And that therefore $-2-x^2$ is always negative?

#### wishmaster

##### Active member
That should be $|x| \sqrt{-2-x^2}$.
Not sure why you would leave out the absolute value symbols.
Oh. Perhaps you do not know what those are?

Let me clarify.
The absolute value of a number x is the magnitude of that number.
One might also say: the number without its sign.
It is written as $|x|$ and can be defined as:
$$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$$

Moving on.
Are you aware that a square like $x^2$ is always either positive or zero, but cannot be negative?
And that therefore $-2-x^2$ is always negative?
yes,i understand the absolute value,so i forgot to put it in the brackets....
But what can i do now with the term under root?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
But what can i do now with the term under root?
Are you aware it is always negative?

#### Jameson

Staff member
To maybe add onto I like Serena's above post, are you familiar with the idea of imaginary numbers? Are you supposed to use the fact that $\sqrt{-1}=i$?

#### wishmaster

##### Active member
To maybe add onto I like Serena's above post, are you familiar with the idea of imaginary numbers? Are you supposed to use the fact that $\sqrt{-1}=i$?
Solution should be without imaginary numbers.
Anyway, i have no time,i have only one hour left to submit the result....
Thank you all for help!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Solution should be without imaginary numbers.
Anyway, i have no time,i have only one hour left to submit the result....
Thank you all for help!
So close...

#### wishmaster

##### Active member
So close...
i will just put that solution is $$\displaystyle \sqrt{-x^4-2x^2}$$, better, not defined for real numbers.