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Completing the Square

shamieh

Active member
Sep 13, 2013
539
How do you complete the square of $6z + 5$ if there isn't a third number. Can someone show me real quick? I've just forgot. First I was thought it would be like 6z + (5/2)^2 + 5 so 6z + 25/4 + 5. Yea I know this is easy, I'm doing it wrong though... In all I just need help completing the square..The problem comes from Calculus II , but I didn't want to ask how to cTs in that Calc forum. Embarassing.

\(\displaystyle \int \frac{6z + 5}{2z + 1}dz\)
 

Pranav

Well-known member
Nov 4, 2013
428
How do you complete the square of $6z + 5$ if there isn't a third number. Can someone show me real quick? I've just forgot. First I was thought it would be like 6z + (5/2)^2 + 5 so 6z + 25/4 + 5. Yea I know this is easy, I'm doing it wrong though... In all I just need help completing the square..The problem comes from Calculus II , but I didn't want to ask how to cTs in that Calc forum. Embarassing.

\(\displaystyle \int \frac{6z + 5}{2z + 1}dz\)
I don't see a reason for completing the square in this case. Its much easier to solve the problem at hand by rewriting it as:
$$\int 3+\frac{2}{2z+1} \,dz$$
 

shamieh

Active member
Sep 13, 2013
539
I don't see a reason for completing the square in this case. Its much easier to solve the problem at hand by rewriting it as:
$$\int 3+\frac{2}{2z+1} \,dz$$
That's what I was thinking...But somehow they got \(\displaystyle \int \frac{3(2z + 1) +2}{2z + 1}\) and I'm just lost on how they get that in the numerator?

anyway you can show me so I can see ?

- - - Updated - - -

Also, 6z/2z = 3 right? so wouldnt you have \(\displaystyle \int 3 + 5\) dx ??
 

Pranav

Well-known member
Nov 4, 2013
428
That's what I was thinking...But somehow they got \(\displaystyle \int \frac{3(2z + 1) +2}{2z + 1}\) and I'm just lost on how they get that in the numerator?
That's basically what I did. You can write $(a+b)/c=a/c+b/c$. So with a=3(2z+1), b=2 and c=2z+1,
$$\int \frac{3(2z+1)+2}{2z+1}=\frac{3(2z+1)}{(2z+1)} + \frac {2} {(2z+1)}=3+\frac{2}{(2z+1)}$$

I hope that helps.
 

shamieh

Active member
Sep 13, 2013
539
The problem is I don't know how to get from \(\displaystyle 6z + 5\) to \(\displaystyle 3(2z + 1) + 2\)
 

Pranav

Well-known member
Nov 4, 2013
428
The problem is I don't know how to get from \(\displaystyle 6z + 5\) to \(\displaystyle 3(2z + 1) + 2\)
$$6z+5=6z+3+2$$
Factor out 3 from the first two terms to get: $3(2z+1)+2$
 

shamieh

Active member
Sep 13, 2013
539
So if I had for example, 6z + 7

I would say 6z + 7 = 6z + 4 + 3
and then get 6z + 7 = 2(3z + 2) + 3?
 

Pranav

Well-known member
Nov 4, 2013
428

shamieh

Active member
Sep 13, 2013
539
Thank you :D