Author Topic: quick complex question (Read 1177 times) Tweet Share

n_sumo23 · « **on:** March 29, 2009, 04:30:24 pm »

hey]
when i solve ix^2 + x + 6i=0 on the calc i get 3i and -2i but when i do it by hand i get -3i and 2i
can some one please figure it out for me/ tell me the right answer with working

cheers

dcc · « **Reply #1 on:** March 29, 2009, 04:37:14 pm »

Quote from: n_sumo23 on March 29, 2009, 04:30:24 pm

$ix^2 + x + 6i = 0$

$\begin{align*} x^2 - ix + 6 &= 0\\ \left(x - \dfrac{i}{2}\right)^2 + \dfrac{1}{4} + 6 &= 0\\ \left(x - \dfrac{i}{2}\right)^2 &= \dfrac{-25}{4}\\ \left(x - \dfrac{i}{2}\right)^2 &= \left(\dfrac{5i}{2}\right)^2\\ x - \dfrac{i}{2} &= \pm \dfrac{5i}{2}\\ x_{1} = 3i,\mbox{ }x_{2} &= -2i\\ \end{align}$

n_sumo23 · « **Reply #2 on:** March 29, 2009, 05:16:13 pm »

thanks do you know how you would do it using the quad formula?

Damo17 · « **Reply #3 on:** March 29, 2009, 05:31:54 pm »

Quote from: n_sumo23 on March 29, 2009, 05:16:13 pm

thanks do you know how you would do it using the quad formula?

you would change $ix^2+x+6i=0$ to $x^2-xi+6=0$ .

$a=1$ , $b=-i$ , $c=6$

$x= \frac{i \pm \sqrt{i^2-4(1)(6)}}{2}$

$= \frac{i \pm \sqrt{-25}}{2}$

$= \frac{i \pm 5i}{2}$

$\therefore$ $x= \frac{i+5i}{2}$ or $x= \frac{i-5i}{2}$

$x= 3i$ or $x=-2i$

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Author Topic: quick complex question (Read 1177 times) Tweet Share

n_sumo23

quick complex question

dcc

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