MHB Solving a Quadratic Binom: Who's Right?

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The discussion revolves around solving the expression $$\sqrt{1-(x^2+1)^2}$$, with participants debating the correct simplification and domain of the resulting function. One participant initially proposed the solution as $$-x^2$$, while another argued for the correct expansion leading to $$\sqrt{-x^4-2x^2}$$. It was clarified that the expression simplifies to $$|x|\sqrt{-2-x^2}$$, indicating that the function is not defined for real numbers due to the negative term under the square root. Ultimately, the consensus is that the composition is not defined in the real number system.
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It was given this term:

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

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

Im wondering who is right now,or how to solve this. Thank you!
 
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wishmaster said:
It was given this term:

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

What are you supposed to do? Find the domain? Simplify?
 
MarkFL said:
What are you supposed to do? Find the domain? Simplify?

Simplify,but i have made corrections,so take a look.
 
wishmaster said:
Simplify,but i have made corrections,so take a look.

Actualy,original question was:

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

Find the composition $$(g\circ f)(x)$$
 
wishmaster said:
Actualy,original question was:

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

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

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

Can you finish?
 
MarkFL said:
$$(g\circ f)(x)=g\left(f(x) \right)=\sqrt{1-\left(x^2+1 \right)^2}$$

Can you finish?

As i wrote:

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

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

Check your expansion of:

$$\left(x^2+1 \right)^2$$
 
MarkFL said:
Check your expansion of:

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

$$x^4+2x^2+1$$ ??

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

but then comes $$\sqrt{-x^4-2x^2}$$ I think i can't simplify that...or?
 
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wishmaster said:
$$x^4+2x^2+1$$ ??

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

but then comes $$\sqrt{-x^4-2x^2}$$ I think i can't 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*}.
 
  • #10
Prove It said:
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:

$$\sqrt{-x^2(x^2+2)}$$
or $$x\sqrt{x^2+2}$$ ??
 
  • #11
wishmaster said:
So then:

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

Yes

or $$x\sqrt{x^2+2}$$ ??

Definitely not! Did you not read my previous post? Also, where have the negatives gone inside your square root?
 
  • #12
Can you show me how?
 
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  • #13
wishmaster said:
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.
 
  • #14
Jameson said:
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:
 
  • #15
Jameson said:
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 I am stuck with this term...
 
  • #16
Pranav said:
I guess you meant $\displaystyle \sqrt{x^2(-2-x^2)}$.

wishmaster said:
I apologize for "u",i know the rules,but sometimes i forget...sorry-1

And now I am 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.
 
  • #17
Pranav said:
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.
 
  • #18
I like Serena said:
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.

$$x\sqrt{-2-x^2}$$
 
  • #19
wishmaster said:
$$x\sqrt{-2-x^2}$$

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?
 
  • #20
I like Serena said:
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?
 
  • #21
wishmaster said:
But what can i do now with the term under root?

Are you aware it is always negative?
 
  • #22
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 $\sqrt{-1}=i$?
 
  • #23
Jameson said:
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!
 
  • #24
wishmaster said:
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...
 
  • #25
I like Serena said:
So close...

i will just put that solution is $$\sqrt{-x^4-2x^2}$$, better, not defined for real numbers.
 
  • #26
wishmaster said:
i will just put that solution is $$\sqrt{-2x^4-2x^2}$$, better, not defined for real numbers.

If you mention the formula, perhaps you want to mention $$\sqrt{-x^4-2x^2}$$ instead.
But yeah, the real answer is that it is not defined for real numbers.
 
  • #27
I like Serena said:
If you mention the formula, perhaps you want to mention $$\sqrt{-x^4-2x^2}$$ instead.
But yeah, the real answer is that it is not defined for real numbers.

Sorry,was my mistake with $$-2x^4$$

I have written that the composition is not defined in real numbers.
Thank you all for the help! Hope i get 100% for homework!
 
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