# Math Help - Integration/ Completing the square...

1. ## 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

2. 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 $x^2 + 6x + 10 = 0$

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, $y=x^2 + 6x + 10$ will never touch the x-axis).

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

what was the original question?

3. ## Re:

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

4. 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,
$x^2+ax$
Then add $\frac{a^2}{4}$ to get,
$x^2+ax+\left(\frac{a}{2}\right)^2 - \frac{a^2}{4}$
Now you can complete the square.
But, note, the front term is positive.

5. 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)