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

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In summary, the conversation was about determining whether the expression √(a^2+2ab-2ac+b^2+2bc+c^2) could be complex or not, given the conditions a>0, b>0, and c>0. The answer was found to be no, and the only way for √x to be complex is if the expression under the square root is negative. The use of a binomial formula was suggested to determine this, which led to the conclusion that the expression must be positive.
  • #1
Tala.S
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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.
 
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  • #2
What is the only way for

[itex]\sqrt{x}[/itex]​

to be complex?

Applied to your case, you know that [itex]a>0,\,b>0,\,c>0[/itex], 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.
 
  • #3
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.
 
  • #4
jfgobin said:
What is the only way for

[itex]\sqrt{x}[/itex]​

to be complex?

Applied to your case, you know that [itex]a>0,\,b>0,\,c>0[/itex], 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 ?
 
  • #5
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 [itex]2ac[/itex]. Do you see two other terms in the expression you could associate with it and use a binomial formula?
 
  • #6
jfgobin said:
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 [itex]2ac[/itex]. 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 ?
 
  • #7
Well, that's [itex] a^2 - 2ac + c^2[/itex] first. That doesn't ring a bell?
 
  • #8
jfgobin said:
Well, that's [itex] a^2 - 2ac + c^2[/itex] 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[itex]\in[/itex]ℝ ?
 
  • #9
Well, that's it:

[itex]a^2 - 2ac + c^2 = \left ( a-c\right)^2[/itex]

Can this become negative if [itex] a,c \in \mathbb{R} [/itex]?
 
  • #10
jfgobin said:
Well, that's it:

[itex]a^2 - 2ac + c^2 = \left ( a-c\right)^2[/itex]

Can this become negative if [itex] a,c \in \mathbb{R} [/itex]?
Yes it can become negative.
 
Last edited:
  • #11
Can it? Really? Can [itex]x^2[/itex] be negative if [itex]x\in\mathbb{R}[/itex]?
 
  • #12
jfgobin said:
Can it? Really? Can [itex]x^2[/itex] be negative if [itex]x\in\mathbb{R}[/itex]?

Oh no of course it can't !
 
  • #13
So, now that you have established that [itex]\left( a-c\right )^2[/itex] cannot be negative, what can you say about [itex]\left( a-c\right )^{2}+2ab+b^2+2bc[/itex]?
 
  • #14
jfgobin said:
So, now that you have established that [itex]\left( a-c\right )^2[/itex] cannot be negative, what can you say about [itex]\left( a-c\right )^{2}+2ab+b^2+2bc[/itex]?

since a>0, b>0, c>0 and [itex]\left( a-c\right )^2[/itex] > 0, the expression [itex]\left( a-c\right )^{2}+2ab+b^2+2bc[/itex] must be positive.
 
  • #15
And now it is time to conclude!

J.
 
  • #16
Yes.

Thank you jfgobin :)
 

1. Can the expression √(a^2+2ab-2ac+b^2+2bc+c^2) be simplified?

Yes, the expression can be simplified by factoring out the common terms. This will result in the simplified expression of √[(a+c)^2+b^2+2b(a+c)].

2. Is it possible for the expression to have complex solutions?

Yes, it is possible for the expression to have complex solutions. This can occur when the terms inside the square root cannot be simplified further and result in imaginary numbers.

3. Can the expression have real solutions?

Yes, the expression can have real solutions. This can occur when the terms inside the square root can be simplified to result in real numbers.

4. Are there any specific conditions for the expression to have complex solutions?

Yes, there are specific conditions for the expression to have complex solutions. The expression will have complex solutions if the discriminant (b^2-4ac) is negative.

5. Can the expression have both real and complex solutions?

Yes, the expression can have both real and complex solutions. This can occur when the discriminant is equal to 0, resulting in a double root, or when the discriminant is positive, resulting in two distinct real roots and two complex roots.

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