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Pious Audio
science forum beginner

Joined: 05 Jun 2006
Posts: 9

Posted: Thu Jun 29, 2006 9:22 pm    Post subject: I don't know why I'm getting the correct answer

I really appreciate the patience on this board; you all don't know how

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?
Gene Wagenbreth
science forum beginner

Joined: 25 May 2005
Posts: 5

 Posted: Thu Jun 29, 2006 9:52 pm    Post subject: Re: I don't know why I'm getting the correct answer 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
amzoti
science forum beginner

Joined: 18 Apr 2006
Posts: 33

Posted: Thu Jun 29, 2006 10:01 pm    Post subject: Re: I don't know why I'm getting the correct answer

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?

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?
Ignacio Larrosa Cañestro1
science forum Guru Wannabe

Joined: 02 May 2005
Posts: 112

Posted: Thu Jun 29, 2006 10:02 pm    Post subject: Re: I don't know why I'm getting the correct answer

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
matt271829-news@yahoo.co.
science forum Guru

Joined: 11 Sep 2005
Posts: 846

Posted: Thu Jun 29, 2006 10:02 pm    Post subject: Re: I don't know why I'm getting the correct answer

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?

(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.
bill11
science forum Guru

Joined: 07 Sep 2005
Posts: 311

Posted: Thu Jun 29, 2006 10:04 pm    Post subject: Re: I don't know why I'm getting the correct answer

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

+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?
William Elliot
science forum Guru

Joined: 24 Mar 2005
Posts: 1906

Posted: Fri Jun 30, 2006 9:53 am    Post subject: Re: I don't know why I'm getting the correct answer

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:

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.

----
The Qurqirish Dragon
science forum Guru Wannabe

Joined: 30 Apr 2005
Posts: 104

Posted: Fri Jun 30, 2006 2:44 pm    Post subject: Re: I don't know why I'm getting the correct answer

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

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),

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