Roots, a+bi form

**MathFailure94** · January 2nd 2011, 05:46 PM

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

Thoughts?

**dwsmith** · January 2nd 2011, 05:50 PM

Quadratic

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

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

**Prove It** · January 2nd 2011, 05:50 PM

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

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

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

Simplify.

**TheCoffeeMachine** · January 2nd 2011, 05:56 PM

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

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

**MathFailure94** · January 2nd 2011, 05:58 PM

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.

**dwsmith** · January 2nd 2011, 06:02 PM

Originally Posted by MathFailure94

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.

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

**Prove It** · January 2nd 2011, 06:03 PM

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

**dwsmith** · January 2nd 2011, 06:07 PM

Originally Posted by Prove It

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

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

**MathFailure94** · January 2nd 2011, 06:17 PM

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!

**dwsmith** · January 2nd 2011, 06:21 PM

Originally Posted by dwsmith

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

$\displaystyle x=\frac{-3}{2}\pm\frac{i}{2}\Rightarrow x=\frac{-3}{2}+\frac{i}{2} \ \mbox{and} \ x=\frac{-3}{2}-\frac{i}{2}$

**Prove It** · January 2nd 2011, 06:53 PM

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

**mr fantastic** · January 2nd 2011, 08:31 PM

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.

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