Question

Solve using any method. -x^2 + 2x - 2 = 0

Answer 1

So what did you try?

Answer 2

Nothing yet.

Answer 3

Well, what method do you think you will need to use?

Answer 4

since it's a quadratic function, what methods could you possibly use to solve for x?

Answer 5

quadratic formula?

Answer 6

With this, that could work well.

Answer 7

okay now what?

Answer 8

Well, know how to put it in that?

Answer 9

determine values for a, b and c

then substitute into the quadratics formula

Answer 10

This is what I am unsure on.

Answer 11

\(\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\)

Answer 12

since it's -x^2 + 2x - 2 = 0

then;
\[a=-1\]
\[b=2\]
\[c=-2\]

Answer 13

substitute these values into the quadratics formula that e.mccormick has stated

Answer 14

And are you up to complex (imaginary) numbers yet? The stuff with i?

Answer 15

\[-2 \pm \sqrt{2^{2-}-4\times=1\times-2}\]

Answer 16

divided by -2

Answer 17

oops 2^2 - 4 x -1 x -2 *

Answer 18

Ummm... OK. I see where the mistake there is. You hit = instad of -.

\(\dfrac{-2\pm \sqrt{2^2-4(-1)(-2)}}{2(-1)}\)

Answer 19

Yeah

Answer 20

Now simplify it. =)

Answer 21

ok

Answer 22

The joy of algebra... and calculus too. You do one easy step. Then you spend the next several minutes simplifying what that did!

Answer 23

\[-2\pm \sqrt{-4}\]

Answer 24

over -2

Answer 25

OK. And that simplifies more. Though that is a good start.

Answer 26

\[(-2\pm2i)/-2\]

Answer 27

Well, that looks like you can make it simpler still. Lot of 2s there.

Answer 28

\[2i\]

Answer 29

?

Answer 30

Nope. But look at what you wrote this way:
\(\dfrac{-2\pm 2i}{-2}\)

What in that cancels?

Answer 31

-2+2

Answer 32

Is that after the cancel or just what cancels?

Answer 33

Not sure...sorry

Answer 34

\(\dfrac{-2\pm 2i}{-2}\implies \dfrac{-2}{-2}\cdot \dfrac{1\mp 1i}{1}\)

That help?

Answer 35

All I did was factor out the -2 on both the top and the bottom. On the top it makes the plus or minus become a minus or plus. That does not really matter because it is still both! But it lets you cancel out the -2 completely.

Answer 36

Okay makes sense.

Answer 37

So, have a final answer now?

Answer 38

basically, the answer is \[1 \pm i \]

Answer 39

or flip the plus/minus sign, it doesnt matter either way

Answer 40

Yep.

Here is how I worked it:

\(\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\)

\(\dfrac{-2\pm \sqrt{2^2-4(-1)(-2)}}{2(-1)}\)

\(\dfrac{-2\pm \sqrt{4-8}}{-2}\)

\(\dfrac{-2\pm \sqrt{-4}}{-2}\)

\(\dfrac{-2\pm \sqrt{4}\sqrt{-1}}{-2}\)

\(\dfrac{-2\pm 2\sqrt{-1}}{-2}\)

\(\dfrac{2\mp 2\sqrt{-1}}{2}\)

\(1\mp \sqrt{-1}\)

\(1\mp i\)

\(\therefore x=(1+i,1-i)\)

Answer 41

Okay yeah! I get it thanks so much!

Answer 42

The quadratic can be a pain in the anatomy, but it is good to remember because sometimes it is just your best choice.

have fun!