1. ## Roots, a+bi form

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

Thoughts?

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

$\displaystyle \displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}\Rightarrow \frac{-6\pm\sqrt{36-4(2)(5)}}{2(2)}=\cdots$

3. $\displaystyle \displaystyle -6x = 2x^2 + 5$

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

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

Simplify.

4. $\displaystyle 2x^2+6x+5 = 0 \Rightarrow 2\left(x+\frac{3}{2}\right)^2+\frac{1}{2} = 0.$

Then it's just one step more to finish it up.

5. Okay so I got it down to

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.

6. Originally Posted by MathFailure94
Okay so I got it down to

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.
$\displaystyle \displaystyle\frac{-6\pm\sqrt{-4}}{4}=\frac{-6\pm 2i}{4}$

7. How does $\displaystyle \displaystyle 6^2 - 4\cdot 2\cdot 5$ equal $\displaystyle \displaystyle 23$?

8. Originally Posted by Prove It
How does $\displaystyle \displaystyle 6^2 - 4\cdot 2\cdot 5$ equal $\displaystyle \displaystyle 23$?
$\displaystyle 36-40\equiv x \ \mbox{(mod 27)}\Rightarrow -4\equiv x \ \mbox{(mod 27)}\Rightarrow -4\equiv 23 \ \mbox{(mod 27)}$

9. 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!

10. Originally Posted by dwsmith
$\displaystyle \displaystyle\frac{-6\pm\sqrt{-4}}{4}=\frac{-6\pm 2i}{4}$
$\displaystyle \displaystyle x=\frac{-3}{2}\pm\frac{i}{2}\Rightarrow x=\frac{-3}{2}+\frac{i}{2} \ \mbox{and} \ x=\frac{-3}{2}-\frac{i}{2}$

11. Originally Posted by dwsmith
$\displaystyle 36-40\equiv x \ \mbox{(mod 27)}\Rightarrow -4\equiv x \ \mbox{(mod 27)}\Rightarrow -4\equiv 23 \ \mbox{(mod 27)}$

Clever! :P

12. Originally Posted by MathFailure94
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!
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.