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Pious Audio
science forum beginner
Joined: 05 Jun 2006
Posts: 9
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 Posted: Thu Jun 29, 2006 9:22 pm Post subject:
I don't know why I'm getting the correct answer
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I really appreciate the patience on this board; you all don't know how
much help you are.
Here's the deal, the direction of a problem I'm working on says:
"solve the following equations by the method of forming a new equation
whose roots are "larger" than those of the original equation by
one-half the coefficient of x."
x^2 + 12x + 9 = 0
so I do as ordered:
y = x + 12/2 = x + 6
x = y - 6
(y - 6)^2 + 12 (y - 6) + 9 = 0
y^2 = 27
y = (27)^(1/2) y = -(27)^(1/2)
x = (27)^(1/2) - 6 x = -(27)^(1/2) - 6
That's all well-and-good, and the book says that I've produced the
correct answer. However, if I plug the durrived value of x into the
original equation, it doesn't equal zero, it equals 9. What's up? |
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Gene Wagenbreth
science forum beginner
Joined: 25 May 2005
Posts: 5
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| Posted: Thu Jun 29, 2006 9:52 pm Post subject:
Re: I don't know why I'm getting the correct answer
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Pious Audio
Your answer is correct. If you plug the derived value in you get zero.
You are making a simple mistake if you get 9.
G |
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amzoti
science forum beginner
Joined: 18 Apr 2006
Posts: 33
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| Posted: Thu Jun 29, 2006 10:01 pm Post subject:
Re: I don't know why I'm getting the correct answer
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Pious Audio wrote:
| Quote: | I really appreciate the patience on this board; you all don't know how
much help you are.
Here's the deal, the direction of a problem I'm working on says:
"solve the following equations by the method of forming a new equation
whose roots are "larger" than those of the original equation by
one-half the coefficient of x."
x^2 + 12x + 9 = 0
so I do as ordered:
y = x + 12/2 = x + 6
x = y - 6
(y - 6)^2 + 12 (y - 6) + 9 = 0
y^2 = 27
y = (27)^(1/2) y = -(27)^(1/2)
x = (27)^(1/2) - 6 x = -(27)^(1/2) - 6
That's all well-and-good, and the book says that I've produced the
correct answer. However, if I plug the durrived value of x into the
original equation, it doesn't equal zero, it equals 9. What's up?
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Actually, you have everything correct:
x^2 = 63 - 36sqrt(3)
12x = - 72 + 36sqrt(3)
So,
x^2 + 12 x + 9 = -9 + 9 = 0
Got it? |
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Ignacio Larrosa Cañestro1
science forum Guru Wannabe
Joined: 02 May 2005
Posts: 112
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| Posted: Thu Jun 29, 2006 10:02 pm Post subject:
Re: I don't know why I'm getting the correct answer
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En el mensaje:1151616175.852506.122060@75g2000cwc.googlegroups.com,
Pious Audio <PiousAudio@gmail.com> escribió:
| Quote: | I really appreciate the patience on this board; you all don't know how
much help you are.
Here's the deal, the direction of a problem I'm working on says:
"solve the following equations by the method of forming a new equation
whose roots are "larger" than those of the original equation by
one-half the coefficient of x."
x^2 + 12x + 9 = 0
so I do as ordered:
y = x + 12/2 = x + 6
x = y - 6
(y - 6)^2 + 12 (y - 6) + 9 = 0
y^2 = 27
y = (27)^(1/2) y = -(27)^(1/2)
x = (27)^(1/2) - 6 x = -(27)^(1/2) - 6
That's all well-and-good, and the book says that I've produced the
correct answer. However, if I plug the durrived value of x into the
original equation, it doesn't equal zero, it equals 9. What's up?
|
Take - 6 + sqrt(27), by example.
(- 6 + sqrt(27))^2 + 12(- 6 + sqrt(27) + 9 =
36 - 12*sqrt(27) + 27 - 72 + 12*sqrt(27) + 9 = 0
Also for - 6 - sqrt(27),
(- 6 - sqrt(27))^2 + 12(- 6 - sqrt(27) + 9 =
36 + 12*sqrt(27) + 27 - 72 - 12*sqrt(27) + 9 = 0
as exected.
You must pay atention when yo raise to the square:
(- 6 + sqrt(27))^2 = (- 6 + sqrt(27))(- 6 + sqrt(27))
= 36 - 6*sqrt(27) - 6*sqrt(27) + sqrt(27)*sqrt(27)
And,
(- 6 - sqrt(27))^2 = (- 6 - sqrt(27))(- 6 - sqrt(27))
= 36 + 6*sqrt(27) + 6*sqrt(27) + sqrt(27)*sqrt(27)
--
Best regards,
Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUITARMAYUSCULAS@mundo-r.com |
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matt271829-news@yahoo.co.
science forum Guru
Joined: 11 Sep 2005
Posts: 846
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| Posted: Thu Jun 29, 2006 10:02 pm Post subject:
Re: I don't know why I'm getting the correct answer
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Pious Audio wrote:
| Quote: | I really appreciate the patience on this board; you all don't know how
much help you are.
Here's the deal, the direction of a problem I'm working on says:
"solve the following equations by the method of forming a new equation
whose roots are "larger" than those of the original equation by
one-half the coefficient of x."
x^2 + 12x + 9 = 0
so I do as ordered:
y = x + 12/2 = x + 6
x = y - 6
(y - 6)^2 + 12 (y - 6) + 9 = 0
y^2 = 27
y = (27)^(1/2) y = -(27)^(1/2)
x = (27)^(1/2) - 6 x = -(27)^(1/2) - 6
That's all well-and-good, and the book says that I've produced the
correct answer. However, if I plug the durrived value of x into the
original equation, it doesn't equal zero, it equals 9. What's up?
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(27^(1/2) - 6)^2 + 12*(27^(1/2) - 6) + 9
= 27 - 12*27^(1/2) + 36 + 12*27^(1/2) - 72 + 9
= 0
and similarly for the other solution, so I think you just made a
calculation error. |
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bill11
science forum Guru
Joined: 07 Sep 2005
Posts: 311
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| Posted: Thu Jun 29, 2006 10:04 pm Post subject:
Re: I don't know why I'm getting the correct answer
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Pious Audio wrote:
| Quote: | I really appreciate the patience on this board; you all don't know how
much help you are.
Here's the deal, the direction of a problem I'm working on says:
"solve the following equations by the method of forming a new equation
whose roots are "larger" than those of the original equation by
one-half the coefficient of x."
x^2 + 12x + 9 = 0
so I do as ordered:
y = x + 12/2 = x + 6
x = y - 6
(y - 6)^2 + 12 (y - 6) + 9 = 0
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+36 - 72 + 9 = - 27; I think.
| Quote: |
y^2 = 27
y = (27)^(1/2) y = -(27)^(1/2)
x = (27)^(1/2) - 6 x = -(27)^(1/2) - 6
That's all well-and-good, and the book says that I've produced the
correct answer. However, if I plug the durrived value of x into the
original equation, it doesn't equal zero, it equals 9. What's up? |
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William Elliot
science forum Guru
Joined: 24 Mar 2005
Posts: 1906
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| Posted: Fri Jun 30, 2006 9:53 am Post subject:
Re: I don't know why I'm getting the correct answer
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From: Pious Audio <PiousAudio@gmail.com>
Newsgroups: sci.math
Subject: I don't know why I'm getting the correct answer
| Quote: | "solve the following equations by the method of forming a new
equation whose roots are "larger" than those of the original
equation by one-half the coefficient of x."
x^2 + 12x + 9 = 0
|
Let r,s be the roots of x^2 + bx + c = 0.
An equation with roots r + b/2, s + b/2 is
(x - r - b/2)(x - s - b/2) = 0
As
0 = x^2 + bx + c = (x - r)(x - s) = x^2 - (r + s)x + rs
one has
r + s = -b; rs = c
To continue, 0
= (x - r - b/2)(x - s - b/2)
= x^2 - (r + s + b)x + rs + (r + s)b/2 + b^2 / 4
= x^2 + c + -bb/2 + b^2 / 4
= x^2 + c - b^2 / 4
has roots r + b/2, s + b/2
c = 9, b = 12, x^2 - 27 = 0
| Quote: | so I do as ordered:
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Don't do that. Understand!
What's happening is the teacher is twisting the discussion about
solving by completing the square into a new fangled problem.
x^2 + bx + c = 0
x^2 + 2(b/2)x + (b/2)^2 = (b/2)^2 - c
(x + b/2)^2 = (b/2)^2 - c
| Quote: | y = (27)^(1/2) y = -(27)^(1/2)
and the book says that I've produced the correct answer.
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The Qurqirish Dragon
science forum Guru Wannabe
Joined: 30 Apr 2005
Posts: 104
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| Posted: Fri Jun 30, 2006 2:44 pm Post subject:
Re: I don't know why I'm getting the correct answer
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Pious Audio wrote:
| Quote: | I really appreciate the patience on this board; you all don't know how
much help you are.
Here's the deal, the direction of a problem I'm working on says:
"solve the following equations by the method of forming a new equation
whose roots are "larger" than those of the original equation by
one-half the coefficient of x."
x^2 + 12x + 9 = 0
so I do as ordered:
y = x + 12/2 = x + 6
x = y - 6
(y - 6)^2 + 12 (y - 6) + 9 = 0
y^2 = 27
y = (27)^(1/2) y = -(27)^(1/2)
x = (27)^(1/2) - 6 x = -(27)^(1/2) - 6
That's all well-and-good, and the book says that I've produced the
correct answer. However, if I plug the durrived value of x into the
original equation, it doesn't equal zero, it equals 9. What's up?
|
As others have stated, you made an error when checking your result.
Also, note that although you cannot solve the problem using other
methods (since you were instructed to do it one particular way), you
can still use other methods to check your answers.
Using the quadratic formula, you find:
(sqrt is the positive square-root function)
x = (-12 +/- sqrt (12^2-4*1*9))/(2*1)
= -6 +/- sqrt(27),
which is exactly what your answers are using the required method. |
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