Can √(a^2+2ab-2ac+b^2+2bc+c^2) Be Complex?

[Tala.S]

Hi

I'm trying to figure out whether this expression √(a^2+2ab-2ac+b^2+2bc+c^2) can be complex or not while a>0, b>0 and c>0.

My answer would be no but I'm not sure.

 

[jfgobin]

What is the only way for

$\sqrt{x}$​

to be complex?

Applied to your case, you know that $a>0,\,b>0,\,c>0$, do you see what term would be the only way to obtain the condition you just found? And is that even possible in this case? Suggestion: once you found the term that can lead to what you seek, use a well known binomial formula.

 

[mfb]

Every real number is a complex number as well. If you are looking for complex numbers with non-zero imaginary part, see the reply above.

 

[Tala.S]

What is the only way for

$\sqrt{x}$​

to be complex?

Applied to your case, you know that $a>0,\,b>0,\,c>0$, do you see what term would be the only way to obtain the condition you just found? And is that even possible in this case? Suggestion: once you found the term that can lead to what you seek, use a well known binomial formula.

x has to be negative ?

The expression can be complex if 2ac>b^2+2bc+c^2+a^2+2ab ?

I can't really see what the term should look like ?

 

[jfgobin]

Tala,

Your on the right track. In order to have an imaginary part - mfb reminded that each real number is also a complex number - the expression under the square root needs to be negative.

You pointed $2ac$. Do you see two other terms in the expression you could associate with it and use a binomial formula?

 

[Tala.S]

Tala,

Your on the right track. In order to have an imaginary part - mfb reminded that each real number is also a complex number - the expression under the square root needs to be negative.

You pointed $2ac$. Do you see two other terms in the expression you could associate with it and use a binomial formula?

well I see the equation a^2+c^2+2ac but I'm not sure why or how to use the binomial formula ?

 

[jfgobin]

Well, that's $a^2 - 2ac + c^2$ first. That doesn't ring a bell?

 

[Tala.S]

Well, that's $a^2 - 2ac + c^2$ first. That doesn't ring a bell?

Not really :S I can't see how I can use a binomial formula ? Is it something like this (a-c)^2=-r,

r$\in$ℝ ?

 

[jfgobin]

Well, that's it:

$a^2 - 2ac + c^2 = \left ( a-c\right)^2$

Can this become negative if $a,c \in \mathbb{R}$?

 

[Tala.S]

Well, that's it:

$a^2 - 2ac + c^2 = \left ( a-c\right)^2$

Can this become negative if $a,c \in \mathbb{R}$?

Yes it can become negative.

 

[jfgobin]

Can it? Really? Can $x^2$ be negative if $x\in\mathbb{R}$?

 

[Tala.S]

Can it? Really? Can $x^2$ be negative if $x\in\mathbb{R}$?

Oh no of course it can't !

 

[jfgobin]

So, now that you have established that $\left( a-c\right )^2$ cannot be negative, what can you say about $\left( a-c\right )^{2}+2ab+b^2+2bc$?

 

[Tala.S]

So, now that you have established that $\left( a-c\right )^2$ cannot be negative, what can you say about $\left( a-c\right )^{2}+2ab+b^2+2bc$?

since a>0, b>0, c>0 and $\left( a-c\right )^2$ > 0, the expression $\left( a-c\right )^{2}+2ab+b^2+2bc$ must be positive.

 

[jfgobin]

And now it is time to conclude!

J.

 

[Tala.S]

Yes.

Thank you jfgobin :)