Completing the square

John Whitaker

Junior Member
Joined
May 9, 2006
Messages
89
The problem:
Solve k^2 + 5k -1 = 0

k^2 + 5k = 1
k^2 + 5k + 25/4 = 4/4 + 25/4
k^2 + 5k + 25/4 = 29/4

The next step my text displays is (k+5/2)^2 = 29/4 without telling me how it arrived at (k+5/2)^2 from the previous k^2 + 5k +25/4 = 29/4

Can you tell me the steps to this part of the problem? Thank you.
John Whitaker
 
The step illustrated in your text seems a little confusing to me as well. Lets see what I can do for you.

With k^2 + 5k + 25/4 = 29/4
you have (k^2 +5/2k) + (5/2k + 25/4) = 29/4
(this is the tricky part, you have to split 5k up into two equal parts, so 5/2k + 5/2k will be your 5k in two equal parts)
factoring: k(k+5/2) + 5/2(k + 5/2) = (k+5/2)^2
[(25/4)/(5/2)= 5/2]

Thus, (k+5/2)^2=29/4

I hope that helps.
 
That's simple factoring, John.
Same as getting (x + 3)^2 from x^2 + 6x + 9.
Notice that 2*3 = 6 and 3^2 = 9;
same with yours: 2*(5/2) = 5 and (5/2)^2 = 25/4.

Glad to see you haven't quit yet :wink:
 
John Whitaker said:
... The next step my text displays is (k+5/2)^2 = 29/4 without telling me how ...


Hi John:

They factored the polynomial. If you use FOIL to expand (k + 5/2)^2, then you will get k^2 + 5k + 25/4.

This is why we complete the square, to obtain a perfect square polynomial that factors.

Cheers,

~ Mark :)

 
Denis,
I'm still struggling... hope this is the last semester. Thanks. (You haven't qiven up either).
John

Gentlemen,
Thank you and I wish you a great holiday.
John
 
I' m sorry... I thought I could figure it out from the responses, but I can't.

(k + 5/2)^2 returns k^2 + 25/4... not k^2 + 5k + 25/4 that I had going in.

Where is the middle term? I'm sorry... I know it's a factoring matter but I need to see the steps that turn k^2 + 5k + 25/4 turn into (k + 5/2)^2

Thanks.
John Whitaker
 
Completing the square II

I' m sorry... I thought I could figure it out from the responses to my first submission, but I can't. The hang up is that k^2 + 5k + 25/4 = 29/4.

The next step is (k + 5/2)^2. I don't see that.

(k + 5/2)^2 returns k^2 + 25/4... not k^2 + 5k + 25/4 that I had going in.

Where is the middle term? I'm sorry... I know it's a factoring matter but I need to see the steps that turn k^2 + 5k + 25/4 turn into (k + 5/2)^2

If you need the whole problem, it is at the top of "Completing the Square"

Thanks.
John Whitaker
 
Re: Completing the square II

John Whitaker said:
I' m sorry... I thought I could figure it out from the responses to my first submission, but I can't. The hang up is that k^2 + 5k + 25/4 = 29/4.

The next step is (k + 5/2)^2. I don't see that.

(k + 5/2)^2 returns k^2 + 25/4 <<< Absolutely NOT

\(\displaystyle (a \, + \, b)^2 \,\)

\(\displaystyle = \, (a \, + \, b)\cdot (a \, + \, b) \,\)

\(\displaystyle = \, a \cdot (a \, + \, b) \, + \, b \cdot (a \, + \, b) \,\)

\(\displaystyle = \, a \cdot a \, + \, a \cdot b \, + \, b \cdot a \, + \, b \cdot b \,\)

\(\displaystyle = \, a^2 \, + \, 2\cdot a \cdot b \, + \, b^2\)



... not k^2 + 5k + 25/4 that I had going in.

Where is the middle term? I'm sorry... I know it's a factoring matter but I need to see the steps that turn k^2 + 5k + 25/4 turn into (k + 5/2)^2

If you need the whole problem, it is at the top of "Completing the Square"

Thanks.
John Whitaker
 
Re: Completing the square II

I thank you for your input, but I'm not smart enough to grasp it. I worked it out substituting the numbers and it did not clarify my problem.

I will appreciate it if, using the numbers I provided, you would show me how k^2 + 5k + 25/4 evolves to (k + 5/2)^2.

Thank you.
John Whitaker
 
Re: Completing the square II

(k + 5/2)^2

= (k + 5/2) * (k + 5/2)

= k * (k + 5/2) + 5/2 * (k + 5/2)

now continue....
 
Re: Completing the square II

Code:
    k + 5/2
    k + 5/2
 ==================
    k^2 + 5k/2
          5k/2 + 25/4
 =====================
    k^2 +  5k  + 25/4
This may help:
put 'em one on top of the other;
multiply top by the bottom k: gives you k^2 + 5k/2
multiply top by the bottom 5/2: gives you 5k/2 + 25/4
then add 'em up.
 


Hi John:

I would like to know whether or not you are familiar with the Distributive Property of Multiplication Over Addition, often called "the distributive property" for short.

If so, then can you use it to multiply the following?

\(\displaystyle k \cdot \left(k + \frac{5}{2} \right) \; = \; ?\)

Gobble, gobble,

~ Mark :)

PS: I heard on the news today that the White House turkey got pardoned. Good grief! He hasn't even left office, yet.
 
Denis, Thanks, I'll try it. Don't give up.

Mark, Thanks. k(k+5/2) = k^2 + 25/4 yes... no...?
WAIT! I think I'm getting it...! I have been forgetting to add the k to the 5/2 part of the problem... I think. I'll try that. If I don't post any more on this... thanks to all who responded and have a great Thanksgiving!
John Whitaker :D
 
Somewhere my problem has been turned around and solutions are being given to the part I understand. I’ll begin again. Please be patient. I can’t move ahead until I understand this

K^2 + 5K -1 = 0 So…
K^2 + 5K = 1

I take ½ of 5 and get 5/2. I square that and get 25/4. So, now my problem reads:

K^2 + 5K + 25/4 = 1 + 25/4 and now my problem begins. My text jumps from there to ( K + 5/2 )^2... skipping the part I need... which is:

From K^2 + 5K + 25/4 how do I get ( K + 5/2 )^2 ?????

I tried multiplying both sides by 4 to get rid of the fraction and went around in a circle. My question is, how does K^2 + 5K + 25/4 evolve to ( K + 5/2 )^2 ???

I know… I seem pretty dense. Well... I am (and that's why I'm back in school.)
Thank you.
John Whitaker
 
For the time being forget your original problem

Take a pencil and paper

Write down the following lines and make sure you understand every step. If you don't - tell us which step you don't understand

\(\displaystyle (a \, + \, b)^2 \,\)

\(\displaystyle = \, (a \, + \, b)\cdot (a \, + \, b) \,\)

\(\displaystyle = \, a \cdot (a \, + \, b) \, + \, b \cdot (a \, + \, b) \,\)

\(\displaystyle = \, a \cdot a \, + \, a \cdot b \, + \, b \cdot a \, + \, b \cdot b \,\)

\(\displaystyle = \, a^2 \, + \, 2\cdot a \cdot b \, + \, b^2\)

Now do you see that:

\(\displaystyle a^2 \, + \, 2\cdot a \cdot b \, + \, b^2 \, = \, (a \, + \, b)^2\)

We cannot move forward till and until you understand this.
 
You can "prove yo yourself" that what Subhotosh showed you works:
(2 + 3)(2 + 3) = (5)(5) = 25
OK:
2(2 + 3) + 3(2 + 3) = 2(5) + 3(5) = 10 + 15 = 25
Even a "struggling student" can't argue with that :wink:

In other words, the contents of the 2nd set of brackets are multiplied by EACH term in the 1st set.

Take (a + b)(b + c):
a(b + c) + b(b + c)
= ab + ac + b^2 + bc
 
Gentlemen,
I do appreciate your efforts... and I'll study what you have shown me, though it still is not as clear as I require (not your fault). I won't post to this again. Thank you.
John Whitaker
 
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