Solving a Quadratic Binom: Who's Right?

[theakdad]

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!

 

[MarkFL]

It was given this term:

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

What are you supposed to do? Find the domain? Simplify?

 

[theakdad]

What are you supposed to do? Find the domain? Simplify?

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

 

[theakdad]

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]

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?

 

[theakdad]

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

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]

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:

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

 

[theakdad]

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 can't simplify that...or?

 

[Prove It]

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

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

but then comes \(\displaystyle \sqrt{-x^4-2x^2}\) I think i 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 $$\displaystyle \begin{align*} \sqrt{x^2} = |x| \end{align*}$$.

 

[theakdad]

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]

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?

 

[theakdad]

Can you show me how?

 

[Jameson]

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.

 

[Saitama]

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

 

[theakdad]

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...

 

[I like Serena]

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 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.

 

[Jameson]

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.

 

[theakdad]

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.

\(\displaystyle x\sqrt{-2-x^2}\)

 

[I like Serena]

\(\displaystyle x\sqrt{-2-x^2}\)

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?

 

[theakdad]

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?

 

[I like Serena]

But what can i do now with the term under root?

Are you aware it is always negative?

 

[Jameson]

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 $\sqrt{-1}=i$?

 

[theakdad]

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!

 

[I like Serena]

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...

 

[theakdad]

So close...

i will just put that solution is \(\displaystyle \sqrt{-x^4-2x^2}\), better, not defined for real numbers.

 

[I like Serena]

i will just put that solution is \(\displaystyle \sqrt{-2x^4-2x^2}\), better, not defined for real numbers.

If you mention the formula, perhaps you want to mention \(\displaystyle \sqrt{-x^4-2x^2}\) instead.
But yeah, the real answer is that it is not defined for real numbers.

 

[theakdad]

If you mention the formula, perhaps you want to mention \(\displaystyle \sqrt{-x^4-2x^2}\) instead.
But yeah, the real answer is that it is not defined for real numbers.

Sorry,was my mistake with \(\displaystyle -2x^4\)

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