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• Jul 23rd 2007, 04:37 PM
Jhevon
Quote:

Originally Posted by BlueStar
You mean y = 9 +- 119i / -2?

no, $\displaystyle y = \frac {9 \mp \sqrt {119}~i}{2}$
• Jul 23rd 2007, 04:38 PM
topsquark
Quote:

Originally Posted by BlueStar
You mean y = 9 +- 119i / -2?

Actually, no.

$\displaystyle y = -x^2 = - \left ( \frac{-9 \pm i \sqrt{119}}{2} \right ) = \frac{9 \mp i \sqrt{119}}{2}$

(There was a typo about the square root sign earlier.)

-Dan
• Jul 23rd 2007, 04:49 PM
Jhevon
Quote:

Originally Posted by topsquark

(There was a typo about the square root sign earlier.)

yes, you are correct. i shall change my posts
• Jul 23rd 2007, 04:57 PM
BlueStar
Okay.

I'll have to be led through on solving this. I cannot brain today, I have the dumb.
• Jul 23rd 2007, 05:03 PM
topsquark
Quote:

Originally Posted by Jhevon
$\displaystyle x^4 + 9x^2 + 50 = 0$

$\displaystyle x^2 = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 1 \cdot 50}}{2 \cdot 1} = \frac{-9 \pm \sqrt{-119}}{2}$

So
$\displaystyle x^2 = \frac{-9 \pm i \sqrt{119}}{2}$

A brief diversion:
$\displaystyle y = -x^2 = - \left ( \frac{-9 \pm i \sqrt{119}}{2} \right ) = \frac{9 \mp i \sqrt{119}}{2}$

Again,
$\displaystyle x^2 = \frac{-9 \pm i \sqrt{119}}{2}$

So
$\displaystyle x =\pm \sqrt{\frac{-9 \pm i \sqrt{119}}{2}}$ <-- The two $\displaystyle \pm$ symbols do not have any relation to each other.

This is technically an unsimplified answer, but in your case I'd leave it like this.

So the solution set is the four points:
The two points
$\displaystyle \left ( \sqrt{\frac{-9 \pm i \sqrt{119}}{2}} , \frac{9 \mp i \sqrt{119}}{2} \right )$
and the two points
$\displaystyle \left (- \sqrt{\frac{-9 \pm i \sqrt{119}}{2}} , \frac{9 \mp i \sqrt{119}}{2} \right )$

-Dan
• Jul 23rd 2007, 05:09 PM
BlueStar
Are those supposed to be the final solutions? Because that seems wrong. It's for a multiple question, and the solution answers only consist of variations of 0 and 5.
• Jul 23rd 2007, 05:12 PM
Jhevon
Quote:

Originally Posted by BlueStar
Are those supposed to be the final solutions? Because that seems wrong. It's for a multiple question, and the solution answers only consist of variations of 0 and 5.

0 and 5 are not solutions to the equations you've posted. the minus sign in from of $\displaystyle (y - 5)^2$ did away with that possibility. 0 would be a solution for x if it was plus
• Jul 23rd 2007, 05:23 PM
BlueStar
Quote:

Originally Posted by Jhevon
x = 0 is NOT a solution (if we have x = 0, that would lead to -25 = 25), and if it was, since y = -x^2, if x = 0, then y = 0, so a solution would be (0,0)

I don't think I wrote the question wrong.
• Jul 23rd 2007, 05:32 PM
Jhevon
Quote:

Originally Posted by BlueStar

I don't think I wrote the question wrong.

i told you, (0,0) is only a solution if you have a + (y - 5)^2. otherwise, there are NO real solutions. if you did not type the question wrong, topsquark's solutions are correct and the solutions you have are wrong. go back to the question and make sure you have the signs right
• Jul 23rd 2007, 05:41 PM
BlueStar
Nope, the signs are right.
• Jul 23rd 2007, 05:54 PM
Jhevon
Quote:

Originally Posted by BlueStar
Nope, the signs are right.

then topsquark's solution is correct
• Jul 23rd 2007, 06:12 PM
BlueStar
Then either the quiz is wrong or topsquark is, or something weird's going on. :confused:
• Jul 23rd 2007, 06:28 PM
Jhevon
Quote:

Originally Posted by BlueStar
Then either the quiz is wrong or topsquark is, or something weird's going on. :confused:

something weird is going on
• Jul 23rd 2007, 08:00 PM
BlueStar
Well, I just selected the answer as (0,0) and it was correct.
• Jul 23rd 2007, 08:08 PM
Jhevon
Quote:

Originally Posted by BlueStar
Well, I just selected the answer as (0,0) and it was correct.

that simply means the minus should have been a plus. good going. it was the best choice of all that were there
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