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anonymous
 2 years ago
Could someone walk me through this question and explain to me what happens. I have been working on it for 7 hours. I don't understand my prof and I might fail the class and I want to cry.
Its solving a quadratic equation that had complex numbers
anonymous
 2 years ago
Could someone walk me through this question and explain to me what happens. I have been working on it for 7 hours. I don't understand my prof and I might fail the class and I want to cry. Its solving a quadratic equation that had complex numbers

This Question is Closed

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0:/ Sorry... I can't help..

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0tried using the quadratic formula yet?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I tried it, i solved it but i got the wrong answers. I tried it over and over again. I looked up videos, still wrong answer

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0\(\large \begin{array}{llll} {\color{red}{ 2}}x^2&{\color{blue}{ i}}x&{\color{green}{ 2+i}}=0\\ \uparrow &\ \uparrow &\quad \uparrow \\ {\color{red}{ a}}&\ {\color{blue}{ b}}&\quad {\color{green}{ c}} \end{array}\\ \quad \\ \bf x=\cfrac{i\pm\sqrt{i^24(2)(2+i)}}{2(2)}\implies x=\cfrac{i\pm\sqrt{18(2+i)}}{2(2)}\\ \quad \\ x=\cfrac{i\pm\sqrt{1+168i}}{2(2)}\implies x=\cfrac{i\pm\sqrt{158i}}{4}\)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0well.. that's.... as far as I can see it going... what does your answer say?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0hmmm I don't see it simplifying ... to that

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1389574709566:dw

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0please explain...never learned that

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1389575164382:dw

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0wait...is this like synthetic division

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1389575421091:dw

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0this is the briotruffini algorithmm, i don't what is synthetic division

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0so in my question the root is 2?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0no the root is one, the first coef in your polynomial is 2

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0how do i know what the root is?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0one, was a easy root to see in that problem

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0oh. so if it had all 4...then it will be 2?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0could u do it step by step and explain what you were doing at each step

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1389575930097:dw