How do I complete the square for $6z + 5$ without a third number?

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Discussion Overview

The discussion revolves around the process of completing the square for the expression $6z + 5$, particularly in the context of an integral from Calculus II. Participants explore different methods and reasoning for manipulating the expression, while also expressing uncertainty about the necessity of completing the square in this scenario.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant asks how to complete the square for $6z + 5$, expressing confusion and seeking help.
  • Another participant suggests that completing the square may not be necessary and proposes rewriting the integral as $\int 3 + \frac{2}{2z + 1} \,dz$ instead.
  • A participant questions how to derive the expression $\int \frac{3(2z + 1) + 2}{2z + 1}$ from $6z + 5$ and seeks clarification.
  • Another participant explains that $6z + 5$ can be rewritten as $3(2z + 1) + 2$ by factoring out 3 from the first two terms.
  • A participant explores a similar example with $6z + 7$ and attempts to apply the same reasoning, leading to a discussion about factoring and rewriting expressions.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of completing the square for the given integral. Some suggest alternative approaches, while others seek clarification on the manipulation of the expressions. The discussion remains unresolved regarding the best method to proceed.

Contextual Notes

Participants express uncertainty about the steps involved in completing the square and the manipulation of expressions, indicating a reliance on specific algebraic techniques that may not be universally agreed upon.

shamieh
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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.

$$\int \frac{6z + 5}{2z + 1}dz$$
 
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shamieh said:
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.

$$\int \frac{6z + 5}{2z + 1}dz$$

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:
$$\int 3+\frac{2}{2z+1} \,dz$$
 
Pranav said:
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 $$\int \frac{3(2z + 1) +2}{2z + 1}$$ 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 wouldn't you have $$\int 3 + 5$$ dx ??
 
shamieh said:
That's what I was thinking...But somehow they got $$\int \frac{3(2z + 1) +2}{2z + 1}$$ 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.
 
The problem is I don't know how to get from $$ 6z + 5$$ to $$3(2z + 1) + 2$$
 
shamieh said:
The problem is I don't know how to get from $$ 6z + 5$$ to $$3(2z + 1) + 2$$

$$6z+5=6z+3+2$$
Factor out 3 from the first two terms to get: $3(2z+1)+2$
 
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?
 
shamieh said:
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?

Yes! :)
 
Thank you :D
 

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