Two Plus Two Older Archives  

Go Back   Two Plus Two Older Archives > General Gambling > Probability
FAQ Community Calendar Today's Posts Search

Reply
 
Thread Tools Display Modes
  #1  
Old 08-22-2005, 07:52 PM
spaminator101 spaminator101 is offline
Senior Member
 
Join Date: Jul 2005
Location: wondering where in the world I left my sweet tea
Posts: 581
Default i need some help

i first posted this in science math and philosophy
they told me to post it here

"so im studying quadratic functions and need help solving quadratic equations by completing the square
im just not getting it"
Reply With Quote
  #2  
Old 08-22-2005, 09:01 PM
BruceZ BruceZ is offline
Senior Member
 
Join Date: Sep 2002
Posts: 1,636
Default Re: i need some help

[ QUOTE ]
i first posted this in science math and philosophy
they told me to post it here

"so im studying quadratic functions and need help solving quadratic equations by completing the square
im just not getting it"

[/ QUOTE ]

Say you have the quadratic 4x^2 + 12x - 7 = 0.

First divide through by 4 to get x^2 by itself:

x^2 + 3x - 7/4 = 0

x^2 + 3x = 7/4

Now the trick is to add something to the left hand side so that we can write it as (x + c)^2. That would be x^2 + 2cx + c^2. We already have x^2 + 2cx where c = 3/2, so we add (3/2)^2 = 9/4 to both sides since x^2 + 3x + 9/4 = (x + 3/2)^2. In general, the magic number that we add is always going to be the coefficient of x divided by 2 and then squared. Remember to make sure the leading coefficient of x^2 is 1 first.

x^2 + 3x <font color="blue">+ 9/4</font> = 7/4 <font color="blue"> + 9/4</font>

(x + 3/2)^2 = 4

x + 3/2 = 2 or x + 3/2 = -2

x = 1/2 or x = -3.5

The whole trick is knowing what to add to each side to make this work. In general, for x^2 + bx + c = 0, the magic term is always (b/2)^2. So you take half the coefficient of x and square it. Don't forget to add it to both sides.

That's how we derive the quadratic formula:

ax^2 + bx + c = 0

x^2 + (b/a)x + c/a = 0

x^2 + (b/a)x = -c/a

x^2 + (b/a)x <font color="blue">+ (b/2a)^2</font> = -c/a <font color="blue">+ (b/2a)^2</font>

(x + b/2a)^2 = -c/a + (b/2a)^2

x + b/2a = +/- sqrt[-c/a + (b/2a)^2]

x = -b/2a +/- sqrt[(-4ac + b^2)/4a^2]

x = [-b +/- sqrt(b^2 - 4ac)] / 2a
Reply With Quote
  #3  
Old 08-22-2005, 10:08 PM
spaminator101 spaminator101 is offline
Senior Member
 
Join Date: Jul 2005
Location: wondering where in the world I left my sweet tea
Posts: 581
Default Re: i need some help

and so basicly you just use the quadratic formula to complete the square

if so this isnt so hard after all
Reply With Quote
  #4  
Old 08-22-2005, 10:13 PM
spaminator101 spaminator101 is offline
Senior Member
 
Join Date: Jul 2005
Location: wondering where in the world I left my sweet tea
Posts: 581
Default Re: i need some help

should i move this part of the discussion to the s.m.p. forum
Reply With Quote
  #5  
Old 08-22-2005, 10:53 PM
BruceZ BruceZ is offline
Senior Member
 
Join Date: Sep 2002
Posts: 1,636
Default Re: i need some help

[ QUOTE ]
and so basicly you just use the quadratic formula to complete the square

if so this isnt so hard after all

[/ QUOTE ]

No, the quadratic formula is another method you can use to solve any quadratic equation, besides completing the square. It's just that we derived the quadratic formula by completing the square for a general quadratic. Then now that we have that formula, we can just use that instead of completing the square manually every time, but your instructor may ask you to solve by completing the square, so then you would have to do it manually as I did in the first example, instead of using the formula.

We can just continue the discussion here.
Reply With Quote
  #6  
Old 08-23-2005, 03:47 PM
spaminator101 spaminator101 is offline
Senior Member
 
Join Date: Jul 2005
Location: wondering where in the world I left my sweet tea
Posts: 581
Default Re: i need some help

ok ive got it now
thanks brucez
Reply With Quote
  #7  
Old 08-24-2005, 11:31 AM
DiceyPlay DiceyPlay is offline
Member
 
Join Date: Jul 2004
Posts: 98
Default A good exercise ...

Use completing the square to derive the quadratic formula.

Start with Ax^2+Bx+c=0

end with x = (-B (+-) SQRT(B^2-4AC))/2A

The algebraic manipulation is a little tricky. This goes a long way in solidifying your algebraic understanding and skills in my opinion.

-DP
Reply With Quote
  #8  
Old 08-24-2005, 04:02 PM
BruceZ BruceZ is offline
Senior Member
 
Join Date: Sep 2002
Posts: 1,636
Default Re: A good exercise ...

[ QUOTE ]
Use completing the square to derive the quadratic formula.

Start with Ax^2+Bx+c=0

end with x = (-B (+-) SQRT(B^2-4AC))/2A

The algebraic manipulation is a little tricky. This goes a long way in solidifying your algebraic understanding and skills in my opinion.

-DP

[/ QUOTE ]

See my first post in this thread.
Reply With Quote
Reply


Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump


All times are GMT -4. The time now is 10:47 PM.


Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2024, vBulletin Solutions Inc.