Question

THIS is the last quadratic equation I need help with tonight, promise

Answer 1

\[y ^{2}+3y-3=0\] is the equation....and so far I have \[-3 + and -\sqrt{21}\over 2\]

Answer 2

Yep.

Answer 3

7 times 3 is 21.. so maybe i wasn't sure if they could simplify

Answer 4

Nope. the square root of 21 is not a nice number.

Answer 5

ok, so what I have is correct. And they are 2 real solutions?

Answer 6

is the thing under the square root a negative number?

Answer 7

no, so if it was -y squared it would be non real?

Answer 8

Yes it would be non-real if a was negative in this case.

Answer 9

or if c was positive and a was positive it would be non-real

Answer 10

You know what we are finding with this equation right?

Answer 11

y?

Answer 12

Imagine we had some equation like:

f(y) = y^2 + 3y -3

Answer 13

And we graphed it. What would it look like?

Answer 14

Where our horizontal axis is y, and the vertical axis is f(y)

Answer 15

http://www.wolframalpha.com/input/?i=graph+f+%3D+y^2+%2B+3y+-+3

Answer 16

When we say the expression equals 0, we are finding where this parabola crosses the horizontal axis.

Answer 17

What is the values for y that make this curve touch the line f(y) = 0

Answer 18

So when we don't get a real solution it means that that curve never touches the f(y)=0 line.

Like this would be the graph if it was -y^2 instead.
http://www.wolframalpha.com/input/?i=graph+f+%3D+-y^2+%2B+3y+-+3

And you can see that since it's opening downward it never crosses the 0 horizontal line.

Answer 19

And this would be the graph if a and c were both positive:

http://www.wolframalpha.com/input/?i=graph+f+%3D+y^2+%2B+3y+%2B+3

Again, no real solutions because the graph doesn't cross f(y)=0.

Answer 20

I hope that helps a bit to put into context what we're doing.

Answer 21

Sorry i was away from this site for a bit. I will check that out, looks helpful! THanks again!