another one :) find the sum and product of the solutions for this equation:
(1+i)x^2 +(3-i)x +(2+4i) = 0

Question

another one :) find the sum and product of the solutions for this equation:
(1+i)x^2 +(3-i)x +(2+4i) = 0

anonymous · Answer

i know that r1r2= c/a   and r1 +r2 = -b/a

anonymous · Answer

By comparing it with:

$$ax^2 + bx + c = 0$$

Can you tell what is a, b and c here ??

anonymous · Answer

okay: a= (1+i) b=(3-i) c= (2+4i)

anonymous · Answer

You have to do same rationalization here..

anonymous · Answer

ah like radicals?

anonymous · Answer

Yes..

anonymous · Answer

thank you, then I can handle it from here :)

anonymous · Answer

Remember:

you will do some mistake in rationalization I guess..

anonymous · Answer

Be careful:

$$i^2 = -1$$

anonymous · Answer

And sorry not you will do..

I mean to say that you can do some mistake..

anonymous · Answer

its okay but here is what i did :)

anonymous · Answer

Yes you can show what you did..

anonymous · Answer

$$\frac{ -(3-i) }{ (1+i) } \times \frac{ (1-i) }{ (1-i) } \rightarrow \frac{ -3+3i+i-i ^{2} }{ 1-i ^{2} }$$

anonymous · Answer

so final is : -1+2i

anonymous · Answer

Yep well done..

anonymous · Answer

thank you for your help @waterineyes !

anonymous · Answer

Do the same for product..

Well Done..

Welcome dear..

anonymous · Answer

yes i did the same for the product and got 3+i =)