Express the roots of the equation-6x = 2x^2 + 5in simplest a+bi form.

Thoughts?

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- Jan 2nd 2011, 05:46 PMMathFailure94Roots, a+bi form
Express the roots of the equation

**-6x = 2x^2 + 5**in simplest a+bi form.

Thoughts? - Jan 2nd 2011, 05:50 PMdwsmith
Quadratic

$\displaystyle 2x^2+6x+5=0$

$\displaystyle \displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}\Rightarrow \frac{-6\pm\sqrt{36-4(2)(5)}}{2(2)}=\cdots$ - Jan 2nd 2011, 05:50 PMProve It
$\displaystyle \displaystyle -6x = 2x^2 + 5$

$\displaystyle \displaystyle 0 = 2x^2 + 6x + 5$.

$\displaystyle \displaystyle x = \frac{-6 \pm \sqrt{6^2 - 4\cdot 2\cdot 5}}{2\cdot 2}$.

Simplify. - Jan 2nd 2011, 05:56 PMTheCoffeeMachine
$\displaystyle 2x^2+6x+5 = 0 \Rightarrow 2\left(x+\frac{3}{2}\right)^2+\frac{1}{2} = 0.$

Then it's just one step more to finish it up. - Jan 2nd 2011, 05:58 PMMathFailure94
Okay so I got it down to

X= -6 +/- radical[23]/4

but how do i proceed from there.

and sorry about the total lack of proper mathematical typing. i have no idea how to type actual radicals, etc. - Jan 2nd 2011, 06:02 PMdwsmith
- Jan 2nd 2011, 06:03 PMProve It
How does $\displaystyle \displaystyle 6^2 - 4\cdot 2\cdot 5$ equal $\displaystyle \displaystyle 23$?

- Jan 2nd 2011, 06:07 PMdwsmith
- Jan 2nd 2011, 06:17 PMMathFailure94
So I'm hazarding a guess that the roots would equal 3i/2 and i/2?

Probably completely incorrect.

I really appreciate your help, though! :) - Jan 2nd 2011, 06:21 PMdwsmith
- Jan 2nd 2011, 06:53 PMProve It
- Jan 2nd 2011, 08:31 PMmr fantastic
You have been told exactly what to do and post #6 even gives you the answer (in unsimplified form). I don't understand why (a) you would still be guessing and (b) you would be guessing wrong. You need to take a lot more care with the basic arithmetic - you have made careless and basic mistakes several times now in this thread.