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**debjit625****Member**- Registered: 2012-07-23
- Posts: 97

Hi all I am having problem with this qudratic equation with signs ,or I am doing something silly ,its all about square root of complex number.

------Eq 1------Eq 2

We can write Eq 2 as

Now we can write Eq 1 as

Is my last equation correct? (I hope so),if yes please solve it using qudratic formula ,as far you can

I have did it but my answer is not matching with correct answer got from other process.The sign is not matching but yes if I write the last equation like this

,my answer matches.

Thanks

Debjit Roy

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The essence of mathematics lies in its freedom - Georg Cantor

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,000

Hi debjit625

Try to substitute y^2=t when you get the quadratic. The solve for t and y will be just the plus/minus square root of it.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**debjit625****Member**- Registered: 2012-07-23
- Posts: 97

No you didnt got my problem ,I know that I am solving for y^2,and after solving for y^2 I have to remove the square,The problem is different just solve it and I will tell it about...

Actually I am solving for this

Debjit Roy

___________________________________________________

The essence of mathematics lies in its freedom - Georg Cantor

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,000

Hi debjit625

Is this what you get?

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**debjit625****Member**- Registered: 2012-07-23
- Posts: 97

Ok before you guys think too much let me tell you where I got this from...as I said its about square root fo a complex number

Let z be a complex number

and its square root be c

Then

So

--- Eq1

--- Eq 2

Now every where they (in books,resource,etc) find x using quadratic formula, and after that they substitute x in Eq2 to find y.But what I am doing is I am trying to find y using quadratic formula ,but my signs(+/-) doesnt matches. I think its math and what process we take it doesnt matters ,the answer will be always same. I must be doing something very very silly...which is not coming into my mind right now but it will come later it always happens with me.

Debjit Roy

___________________________________________________

The essence of mathematics lies in its freedom - Georg Cantor

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,000

Hi debjit625

What do you get as the answer and what does the book get as the answer?

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**debjit625****Member**- Registered: 2012-07-23
- Posts: 97

anonimnystefy yes what you got I also got that using that equation,but things changes when I use

The correct answer will be

And thanks for reply

*Last edited by debjit625 (2012-07-29 20:58:11)*

___________________________________________________

The essence of mathematics lies in its freedom - Georg Cantor

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,000

Hi debjit625

That is my solution with the plus/minus sign inside the root taken as +. I am sure taking the minus sign would work as well.

What did you say the correct solution should be?

*Last edited by anonimnystefy (2012-07-29 21:01:51)*

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**debjit625****Member**- Registered: 2012-07-23
- Posts: 97

anonimnystefy wrote:

That is my solution with the plus/minus sign inside the root taken as +. I am sure taking the minus sign would work as well.

No the minus will not work in this case , as we are finding y^2 and y^2 cant be negative.

But thats not the big problem of mine...my problem is that I can't get to the answer using this equation

And the correct answer for x and y in my book and every where else is

Another thing I want to say,when I write the equation

to this,i.e.. moving the equation from one side to another

I get correct answer.Is here I am doing something wrong?

*Last edited by debjit625 (2012-07-29 21:27:52)*

___________________________________________________

The essence of mathematics lies in its freedom - Georg Cantor

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,000

Moving everything shouldn't change anything.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**debjit625****Member**- Registered: 2012-07-23
- Posts: 97

Ok ,I was going to ask to you to show the steps but suddenly I realized that I have got it...

Actually I answered my question in post #9

Thanks for your time bro, "talking to you made my crazy mind to work"

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The essence of mathematics lies in its freedom - Georg Cantor

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,000

You're welcome. But I am sure you are not crazy.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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