Solving Quadratic Systems

**Jhevon** · July 23rd 2007, 05:37 PM

Originally Posted by BlueStar

You mean y = 9 +- 119i / -2?

no, $y = \frac {9 \mp \sqrt {119}~i}{2}$

**topsquark** · July 23rd 2007, 05:38 PM

Originally Posted by BlueStar

You mean y = 9 +- 119i / -2?

Actually, no.

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

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

-Dan

**Jhevon** · July 23rd 2007, 05:49 PM

Originally Posted by topsquark

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

yes, you are correct. i shall change my posts

**BlueStar** · July 23rd 2007, 05:57 PM

Okay.

I'll have to be led through on solving this. I cannot brain today, I have the dumb.

**topsquark** · July 23rd 2007, 06:03 PM

Originally Posted by Jhevon

$x^4 + 9x^2 + 50 = 0$

The quadratic formula says that
$x^2 = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 1 \cdot 50}}{2 \cdot 1} = \frac{-9 \pm \sqrt{-119}}{2}$

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

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

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

So
$x =\pm \sqrt{\frac{-9 \pm i \sqrt{119}}{2}}$ <-- The two $\pm$ 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
$\left ( \sqrt{\frac{-9 \pm i \sqrt{119}}{2}} , \frac{9 \mp i \sqrt{119}}{2} \right )$
and the two points
$\left (- \sqrt{\frac{-9 \pm i \sqrt{119}}{2}} , \frac{9 \mp i \sqrt{119}}{2} \right )$

-Dan

**BlueStar** · July 23rd 2007, 06:09 PM

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.

**Jhevon** · July 23rd 2007, 06:12 PM

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 $(y - 5)^2$ did away with that possibility. 0 would be a solution for x if it was plus

**BlueStar** · July 23rd 2007, 06:23 PM

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)

What about this? That's the only possible solution that's on the multiple choice.

I don't think I wrote the question wrong.

**Jhevon** · July 23rd 2007, 06:32 PM

Originally Posted by BlueStar

What about this? That's the only possible solution that's on the multiple choice.

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

**BlueStar** · July 23rd 2007, 06:41 PM

Nope, the signs are right.

**Jhevon** · July 23rd 2007, 06:54 PM

Originally Posted by BlueStar

Nope, the signs are right.

then topsquark's solution is correct

**BlueStar** · July 23rd 2007, 07:12 PM

Then either the quiz is wrong or topsquark is, or something weird's going on.

**Jhevon** · July 23rd 2007, 07:28 PM

Originally Posted by BlueStar

Then either the quiz is wrong or topsquark is, or something weird's going on.

something weird is going on

**BlueStar** · July 23rd 2007, 09:00 PM

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

**Jhevon** · July 23rd 2007, 09:08 PM

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

Math Help - Solving Quadratic Systems

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