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[SOLVED] Term under root

wishmaster

Active member
Oct 11, 2013
211
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!
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

wishmaster

Active member
Oct 11, 2013
211

wishmaster

Active member
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Oct 11, 2013
211
\(\displaystyle (g\circ f)(x)=g\left(f(x) \right)=\sqrt{1-\left(x^2+1 \right)^2}\)

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

Administrator
Staff member
Feb 24, 2012
13,775
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?
Check your expansion of:

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

wishmaster

Active member
Oct 11, 2013
211
Check your expansion of:

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

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?
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
\(\displaystyle x^4+2x^2+1\) ??

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 [tex]\displaystyle \begin{align*} x^2 \end{align*}[/tex] and remember that [tex]\displaystyle \begin{align*} \sqrt{x^2} = |x| \end{align*}[/tex].
 

wishmaster

Active member
Oct 11, 2013
211
You can simplify it if you take out a factor of [tex]\displaystyle \begin{align*} x^2 \end{align*}[/tex] and remember that [tex]\displaystyle \begin{align*} \sqrt{x^2} = |x| \end{align*}[/tex].
So then:

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

Prove It

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MHB Math Helper
Jan 26, 2012
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wishmaster

Active member
Oct 11, 2013
211
Can you show me how?
 
Last edited:

Jameson

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Jan 26, 2012
4,040
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
Nov 4, 2013
428
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 $\displaystyle \sqrt{x^2(-2-x^2)}$. :rolleyes:
 

wishmaster

Active member
Oct 11, 2013
211
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
Mar 5, 2012
8,774
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

Administrator
Staff member
Jan 26, 2012
4,040
I guess you meant $\displaystyle \sqrt{x^2(-2-x^2)}$. :rolleyes:
Doh! You're quite right. That'll teach me to do math just after waking up.
 

wishmaster

Active member
Oct 11, 2013
211
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
Mar 5, 2012
8,774
\(\displaystyle x\sqrt{-2-x^2}\)
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
Oct 11, 2013
211
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
Mar 5, 2012
8,774

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
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
Oct 11, 2013
211
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
Mar 5, 2012
8,774
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
Oct 11, 2013
211