- #1

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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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- Thread starter Tala.S
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- #1

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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.

- #2

- 90

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[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

mfb

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- #4

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[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

- 90

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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

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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

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

- #8

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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

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[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

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[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.

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- #11

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Can it? Really? Can [itex]x^2[/itex] be negative if [itex]x\in\mathbb{R}[/itex]?

- #12

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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

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- #14

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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

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And now it is time to conclude!

J.

J.

- #16

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Yes.

Thank you jfgobin :)

Thank you jfgobin :)

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