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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

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

- katieb

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

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

- anonymous

Yep.

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

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

- anonymous

is the thing under the square root a negative number?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

You know what we are finding with this equation right?

- anonymous

y?

- anonymous

Imagine we had some equation like:
f(y) = y^2 + 3y -3

- anonymous

And we graphed it. What would it look like?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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.

- anonymous

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.