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

  • MHB
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In summary, to complete the square of an expression like 6z + 5, you can factor out a common factor of 3 to get 3(2z + 1) + 2, which can then be rewritten as 3 + 2/(2z + 1). This can then be integrated easily, as shown in the conversation above.
  • #1
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.

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

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

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:
$$\int 3+\frac{2}{2z+1} \,dz$$
 
  • #3
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 \(\displaystyle \int \frac{3(2z + 1) +2}{2z + 1}\) 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 wouldn't you have \(\displaystyle \int 3 + 5\) dx ??
 
  • #4
shamieh said:
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.
 
  • #5
The problem is I don't know how to get from \(\displaystyle 6z + 5\) to \(\displaystyle 3(2z + 1) + 2\)
 
  • #6
shamieh said:
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: $3(2z+1)+2$
 
  • #7
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?
 
  • #8
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! :)
 
  • #9
Thank you :D
 

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

What is the purpose of completing the square?

The purpose of completing the square is to solve quadratic equations that cannot be easily factored. It allows us to find the vertex, or maximum/minimum point, of a parabola and to rewrite quadratic equations in a standard form that is useful for graphing and solving.

How do you complete the square?

To complete the square, take half of the coefficient of the x term and square it, then add that value to both sides of the equation. This will create a perfect square trinomial on the left side of the equation. You can then factor it to find the solutions.

Can completing the square be used for all quadratic equations?

Yes, completing the square can be used for all quadratic equations. However, it is typically only used when the quadratic equation cannot be easily factored.

Why is completing the square useful?

Completing the square is useful because it allows us to solve quadratic equations that cannot be easily factored. It also helps us find the vertex of a parabola, which can be used to graph the quadratic equation. In addition, completing the square can help us rewrite quadratic equations in a standard form that is useful for solving and graphing.

Are there any shortcuts for completing the square?

Yes, there are some shortcuts for completing the square. One common shortcut is the "b/2" method, where you take half of the coefficient of the x term and square it to add to both sides of the equation. Another shortcut is the "completing the square" formula, which is -b/2±√(b/2)^2-c, where b represents the coefficient of the x term and c represents the constant term.

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