# Math Help - Quadratic equations

I am working with some quadratic equations(from my notes) and I am having trouble understanding what is going on.

For one I have....
1.
3y^2-6y-2=0

then it goes to

2.
y=[6+√(36+24)]/6 and I am not sure what happened there. EDIT:woops forgot about the quadratic formula :P

then

3.
y=6+√(4*15) ->I don't know what happened to the 6 that was on the bottom.

4.
y=(3+√15)/3 ->where did the 4 go? where did these 3's come from?

5.
y=1+[3/(3+√15)] ->still confused

6.(3+√15+3)/(3+√15) ->I actually know what happened here

7.
y=[(6+√15)(3-√15)]/[(3+√15)(3-√15)] ->What??!?! No clue how we got from step 6 to 7.

8.
y=(18+3√15-6√15-15)/(9-15) ->I can see what happened here

9.y=3-3√15 ->what happened to the 9-15 on the bottom

ps. I am pretty sure I copied the notes down correctly but maybe I am wrong.

Originally Posted by ehpoc
I am working with some quadratic equations(from my notes) and I am having trouble understanding what is going on.

For one I have....
1.
3y^2-6y-2=0

then it goes to

2.
y=[6+√(36+24)]/6 and I am not sure what happened there. EDIT:woops forgot about the quadratic formula :P
Yep!
Note that this is only one of the 2 solutions (the "+" one).

Originally Posted by ehpoc
then

3.
y=6+√(4*15) ->I don't know what happened to the 6 that was on the bottom.
Yep, someone dropped the 6 by accident.

Originally Posted by ehpoc
4.
y=(3+√15)/3 ->where did the 4 go? where did these 3's come from?
The 4 was factored out of the square root.
√(4*15) = √4 √15 = 2 √15

At the same time both nominator and denomitor were divided by 2.
As you can see the lost 6 in the denominator magically reappeared here.

Originally Posted by ehpoc
5.
y=1+[3/(3+√15)] ->still confused
I'm afraid his step is incorrect.
So what comes after is also incorrect.
y=3-3√15 is not a solution to the quadratic equation.

Originally Posted by ehpoc
6.(3+√15+3)/(3+√15) ->I actually know what happened here

7.
y=[(6+√15)(3-√15)]/[(3+√15)(3-√15)] ->What??!?! No clue how we got from step 6 to 7.

8.
y=(18+3√15-6√15-15)/(9-15) ->I can see what happened here

9.y=3-3√15 ->what happened to the 9-15 on the bottom

ps. I am pretty sure I copied the notes down correctly but maybe I am wrong.

Here is what I did, tell me if this is still technically correct.

3x^2-6x-2=0

I divided by 3

x^2-2x-2/3=0

x=[2√(4+8/3)]/2

x=√20/8

You appear to have left out part of the quadratic formula.

The solution of
$ax^2+bx+c=0$

is
$x={-b \pm \sqrt{b^2 - 4ac} \over 2a}$
What did you do with (-b)?

It is there....

the 2?

but the 2 on top and 2 on bottom cancel out right?

Aha!
In that case, what did you do with the $\pm$?

Oh sorry....I am using this as applied to solving continued fractions, and thus in a continued fraction it is positive.

But that aside does it look correct?

There should be a plus (or a minus) between the 2 and the square root.
Effectively this means that the 2 in the nominator does not simply cancel against the 2 in the denominator.

Perhaps you can add a plus in between and redo you calculation?

won't I end up with the exact same answer but with a plus/minus symbol infront?

Huh?

With a plus in between, it's an addition.
Without anything in between, it's an implied multiplication.

How would that be able to yield the same answer?

woops ya never thought about that.

so it would be in this case (because it is a continued fraction it is plus)

x=(2+√20/8)/2

Is this correct? Is the √(20/8) correct?

Originally Posted by ehpoc
woops ya never thought about that.

so it would be in this case (because it is a continued fraction it is plus)

x=(2+√20/8)/2

Is this correct? Is the √(20/8) correct?
First off, I don't understand your reference to a continued fraction.
That is not what this is.

But I'm afraid √(20/8) is not correct.
This comes down to adding fractions:

$4 + {8 \over 3} = {4 \cdot 3 \over 3} + {8 \over 3} = {4 \cdot 3 + 8 \over 3}$

This quadratic equation is being using in solving for x in a continued fraction. If you don't know what that is it does not really matter. All it means is that the plus/minus changes to a plus.

Anyways that was a typo

I meant

x=(2+√20/3)/2