# Roots, a+bi form

• January 2nd 2011, 06:46 PM
MathFailure94
Roots, a+bi form
Express the roots of the equation -6x = 2x^2 + 5 in simplest a+bi form.

Thoughts?
• January 2nd 2011, 06:50 PM
dwsmith

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

$\displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}\Rightarrow \frac{-6\pm\sqrt{36-4(2)(5)}}{2(2)}=\cdots$
• January 2nd 2011, 06:50 PM
Prove It
$\displaystyle -6x = 2x^2 + 5$

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

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

Simplify.
• January 2nd 2011, 06:56 PM
TheCoffeeMachine
$2x^2+6x+5 = 0 \Rightarrow 2\left(x+\frac{3}{2}\right)^2+\frac{1}{2} = 0.$

Then it's just one step more to finish it up.
• January 2nd 2011, 06:58 PM
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.
• January 2nd 2011, 07:02 PM
dwsmith
Quote:

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.

$\displaystyle\frac{-6\pm\sqrt{-4}}{4}=\frac{-6\pm 2i}{4}$
• January 2nd 2011, 07:03 PM
Prove It
How does $\displaystyle 6^2 - 4\cdot 2\cdot 5$ equal $\displaystyle 23$?
• January 2nd 2011, 07:07 PM
dwsmith
Quote:

Originally Posted by Prove It
How does $\displaystyle 6^2 - 4\cdot 2\cdot 5$ equal $\displaystyle 23$?

$36-40\equiv x \ \mbox{(mod 27)}\Rightarrow -4\equiv x \ \mbox{(mod 27)}\Rightarrow -4\equiv 23 \ \mbox{(mod 27)}$

:)
• January 2nd 2011, 07:17 PM
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! :)
• January 2nd 2011, 07:21 PM
dwsmith
Quote:

Originally Posted by dwsmith
$\displaystyle\frac{-6\pm\sqrt{-4}}{4}=\frac{-6\pm 2i}{4}$

$\displaystyle x=\frac{-3}{2}\pm\frac{i}{2}\Rightarrow x=\frac{-3}{2}+\frac{i}{2} \ \mbox{and} \ x=\frac{-3}{2}-\frac{i}{2}$
• January 2nd 2011, 07:53 PM
Prove It
Quote:

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

:)

Clever! :P
• January 2nd 2011, 09:31 PM
mr fantastic
Quote:

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.