Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**debjit625****Member**- Registered: 2012-07-23
- Posts: 89

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

___________________________________________________

The essence of mathematics lies in its freedom - Georg Cantor

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,804

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

Offline

**debjit625****Member**- Registered: 2012-07-23
- Posts: 89

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

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,804

Hi debjit625

Is this what you get?

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

Offline

**debjit625****Member**- Registered: 2012-07-23
- Posts: 89

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

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,804

Hi debjit625

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

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

Offline

**debjit625****Member**- Registered: 2012-07-23
- Posts: 89

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

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,804

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)*

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

Offline

**debjit625****Member**- Registered: 2012-07-23
- Posts: 89

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

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,804

Moving everything shouldn't change anything.

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

Offline

**debjit625****Member**- Registered: 2012-07-23
- Posts: 89

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"

___________________________________________________

The essence of mathematics lies in its freedom - Georg Cantor

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,804

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

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

Offline

Pages: **1**