1. ## Impossible equation?

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.

2. Look at the graph. Do the two intersect?

3. 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!

4. 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.

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

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

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

$i = \sqrt{-1}$

6. 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 $\frac {-2}{3}$