# Integration/ Completing the square...

• May 30th 2007, 12:10 PM
qbkr21
Integration/ Completing the square...
Hi guys sorry I haven't been online much I've been very busy with SS1...

Quick Question:

If I am completing the square and I get the point where I am about the square both sides, but the side on the right is a negative number what am I to do. By the rule you cannot take the square root of a negative number. So after examining this situation I have come to the conclusion that I might should move the # back over to the left hand side.

Here is my case

(x+3)^2=-1

See my Dilemma...

Thanks,

-qbkr21
• May 30th 2007, 12:16 PM
Jhevon
Quote:

Originally Posted by qbkr21
Hi guys sorry I haven't been online much I've been very busy with SS1...

Quick Question:

If I am completing the square and I get the point where I am about the square both sides, but the side on the right is a negative number what am I to do. By the rule you cannot take the square root of a negative number. So after examining this situation I have come to the conclusion that I might should move the # back over to the right hand side.

Here is my case

(x+3)^2=-1

See my Dilemma...

Thanks,

-qbkr21

SS1?

i assume the original eqution was $\displaystyle x^2 + 6x + 10 = 0$

if so you are correct. you would get roots with imaginary numbers, since the above is never zero (you can see this by plotting the graph, $\displaystyle y=x^2 + 6x + 10$ will never touch the x-axis).

we define $\displaystyle \sqrt {-1}$ as $\displaystyle i$, which is the basis for defining the imaginary part of a complex number. the above equation has complex roots

what was the original question?
• May 30th 2007, 12:22 PM
qbkr21
Re:
Quote:

Originally Posted by Jhevon
SS1?

Summer School First Session...

Yea you were right on the money it was (1)/(x^2+6x+10). My professor actually kept the (x+3)^2+1. This way I could integrate the function with partial frac's. One other question.

I am wondering why on this example

(6x-x^2) he went ahead and factored out the negative to make it

-(x^2-6x) Why did he do this? Is there rule with completing the square that you cannot have negatives? If so which can't be negative...

Thanks,

qbkr21:)
• May 30th 2007, 12:24 PM
ThePerfectHacker
Quote:

Originally Posted by qbkr21

-(x^2-6x) Why did he do this? Is there rule with completing the square that you cannot have negatives? If so which can't be negative...

Because the general form is,
$\displaystyle x^2+ax$
Then add $\displaystyle \frac{a^2}{4}$ to get,
$\displaystyle x^2+ax+\left(\frac{a}{2}\right)^2 - \frac{a^2}{4}$
Now you can complete the square.
But, note, the front term is positive.
• May 30th 2007, 12:25 PM
Jhevon
Quote:

Originally Posted by qbkr21
Summer School First Session...

Yea you were right on the money it was (1)/(x^2+6x+10). My professor actually kept the (x+3)^2+1. This way I could integrate the function with partial frac's. One other question.

I am wondering why on this example

(6x-x^2) he went ahead and factored out the negative to make it

-(x^2-6x) Why did he do this? Is there rule with completing the square that you cannot have negatives? If so which can't be negative...

Thanks,

qbkr21:)

i'll think about why he did that, but i wouldn't. after completing the square, i would use substitution, let u = x + 3, then the answer would be arctan(u)