Impossible equation?

**Sodapop** · April 12th 2009, 11:37 AM

I've been working on it for a few days, and I can't seem to find a possible answer. My maths teacher told me it was not impossible (and she's not the joking kind).

$x^2 + y^2 = 4$
$y= -x+3$

So let me show you what I tried...

$x^2 + y^2 = 4$
$x^2 + (-x+3)^2 = 4$
$x^2 + (-x+3)(-x+3) = 4$
$x^2 + (x^2-3x-3x+9) = 4$
$2x^2 -6x + 5 = 0$

Then...

$\frac {-b \pm \sqrt {b^2-4ac}}{2a}$ so $\frac {6 \pm \sqrt {36-40}}{4}$ so $\frac {6 \pm \sqrt {-4}}{4}$

And $\sqrt {-4}$ is obviously the problem here.

Any ideas? Thanks.

**Plato** · April 12th 2009, 11:49 AM

Look at the graph. Do the two intersect?

**Sodapop** · April 12th 2009, 12:02 PM

Originally Posted by Plato

Look at the graph. Do the two intersect?

Ohhh, thanks! Tracing it on a graph is always easier, now I get it =p Well, time to bash my teacher now!

**sammy28** · April 12th 2009, 01:34 PM

we are currently looking at uses of the discriminant in school. according to the text book there are no real roots but 2 complex roots when its < 0.

can these roots be shown. haven't got to this part yet at school.

**metlx** · April 12th 2009, 02:00 PM

so

$x_{1,2} =\frac{6 \pm \sqrt{-4}}{4} = \frac {6 \pm 2i}{4}$

$x_1 = \frac {3}{2} + \frac {i}{2}$

$x_2 = \frac{3}{2} - \frac {i}{2}$

$i = \sqrt{-1}$

**Sodapop** · April 12th 2009, 02:24 PM

Originally Posted by sammy28

we are currently looking at uses of the discriminant in school. according to the text book there are no real roots but 2 complex roots when its < 0.

can these roots be shown. haven't got to this part yet at school.

Yes, two answers are found when it's < 0. Here is an example:

$a=3$
$b=5$
$c=2$

$\frac {-b \pm \sqrt {b^2-4ac}}{2a}$

$\frac {-5 \pm \sqrt {5^2-4(3)(2)}}{2(3)}$

$\frac {-5 \pm \sqrt {25-24}}{6}$

$\frac {-5 \pm \sqrt {1}}{6}$

$\frac {-5 \pm {1}}{6}$

So here are both of the answers:

$\frac {-5 - {1}}{6}$ and $\frac {-5 +{1}}{6}$

Simplified, you get:

$-1$ and $\frac {-2}{3}$

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