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> Negative Square Roots by PJS
PJS
post Jul 07, 2010, 10:44 AM
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sqrt -16 = x
-sqrt 16 = - x
-4 = -x
-(4)^2 = -(x)^2
-16 = -16
sqrt -16 = sqrt -x^2 = sqrt (-x * -x)
sqrt -16 = -x, now multiply both sides by negative 1 again
-sqrt 16 = x
-4 = x
sqrt -16 = -4

P.j.S
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Homegamek
post Jul 07, 2010, 11:52 AM
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QUOTE(PJS @ Jul 07, 2010, 07:44 PM) *

sqrt -16 = x
-sqrt 16 = - x
-4 = -x
-(4)^2 = -(x)^2
-16 = -16
sqrt -16 = sqrt -x^2 = sqrt (-x * -x)
sqrt -16 = -x, now multiply both sides by negative 1 again
-sqrt 16 = x
-4 = x
sqrt -16 = -4

P.j.S


I wonder you discard the complex numbers??? Consider what you wrote in frames of conventional rules of the mathematics, I do not understand what you want to say by your writings. Your transition between lines 7 and 8 contains ERROR: see point J.

A)
Input: Solve[Sqrt[-16] == x]
Output: x= 4 * i , where i is the imaginary number (here and below the sign "*" is used as multiplication sign).

cool.gif - Sqrt[16]= - x <=> (-1)* Sqrt[16]= (-1)* x <=> Sqrt[16]= x
Input: Solve[Sqrt[16] == x]
Output: x=4

C) -4 = -x <=> (-1)*4 =(-1)*x

((-1)*4)^2 =((-1)*x )^2 =>
((-1)^2)*(4^2) =((-1)^2)*(x ^2 ) , then
as (-1)^2=1, we have
4^2=x^2

Input: Solve[x^2=4^2]
Output: x1=-4, x2=4 , so the equation has 2 solutions.

D) -16 = -16 = (-1)* (4^2)=(-1)* ((-4)^2)

E1) Sqrt[-16] = Sqrt[(-x)^2] <==> Sqrt[-16] = Sqrt[x^2]

Input: Solve[Sqrt[-16] == Sqrt[(-x)^2]]
Output: x1=-4*i, x2=4*i , where i is the complex number, so the equation has 2 solutions.


E2)Input: Solve[Sqrt[-16] == Sqrt[-x^2]]
Output: x1=-4, x2=4 , so the equation has 2 solutions.

F) Input: Solve[Sqrt[-16] == -x]
Output: x=-4*i , where i is the complex number

J) If you multiply both sides of Sqrt[-16] = -x by (-1) you will get -Sqrt[-16] = x and not -Sqrt[16] = x as you mention.

Input: Solve[-Sqrt[-16] == x]
Output: x= -4*i
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PJS
post Jul 08, 2010, 08:50 AM
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QUOTE(Homegamek @ Jul 07, 2010, 11:52 AM) *

QUOTE(PJS @ Jul 07, 2010, 07:44 PM) *

sqrt -16 = x
-sqrt 16 = - x
-4 = -x
-(4)^2 = -(x)^2
-16 = -16
sqrt -16 = sqrt -x^2 = sqrt (-x * -x)
sqrt -16 = -x, now multiply both sides by negative 1 again
-sqrt 16 = x
-4 = x
sqrt -16 = -4

P.j.S


I wonder you discard the complex numbers??? Consider what you wrote in frames of conventional rules of the mathematics, I do not understand what you want to say by your writings. Your transition between lines 7 and 8 contains ERROR: see point J.

A)
Input: Solve[Sqrt[-16] == x]
Output: x= 4 * i , where i is the imaginary number (here and below the sign "*" is used as multiplication sign).

cool.gif - Sqrt[16]= - x <=> (-1)* Sqrt[16]= (-1)* x <=> Sqrt[16]= x
Input: Solve[Sqrt[16] == x]
Output: x=4

C) -4 = -x <=> (-1)*4 =(-1)*x

((-1)*4)^2 =((-1)*x )^2 =>
((-1)^2)*(4^2) =((-1)^2)*(x ^2 ) , then
as (-1)^2=1, we have
4^2=x^2

Input: Solve[x^2=4^2]
Output: x1=-4, x2=4 , so the equation has 2 solutions.

D) -16 = -16 = (-1)* (4^2)=(-1)* ((-4)^2)

E1) Sqrt[-16] = Sqrt[(-x)^2] <==> Sqrt[-16] = Sqrt[x^2]

Input: Solve[Sqrt[-16] == Sqrt[(-x)^2]]
Output: x1=-4*i, x2=4*i , where i is the complex number, so the equation has 2 solutions.


E2)Input: Solve[Sqrt[-16] == Sqrt[-x^2]]
Output: x1=-4, x2=4 , so the equation has 2 solutions.

F) Input: Solve[Sqrt[-16] == -x]
Output: x=-4*i , where i is the complex number

J) If you multiply both sides of Sqrt[-16] = -x by (-1) you will get -Sqrt[-16] = x and not -Sqrt[16] = x as you mention.

Input: Solve[-Sqrt[-16] == x]
Output: x= -4*i

x^3 - 2x^2 = 1 proves that imaginary numbers are in error. See x^2 + 1 = 0. x^2 = -1.
x^3 - 2(-1) = 1
x^3 + 2 = 1
x^3 = 1 - 2
x^3 = -1
x = -1 , x^2 = 1 So x^2 = -1 and 1 at the same time. Another arrangement for negative square roots may be needed.

P.j.S
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PJS
post Jul 08, 2010, 09:21 AM
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QUOTE(PJS @ Jul 07, 2010, 10:44 AM) *

sqrt -16 = x
-sqrt 16 = - x
-4 = -x
-(4)^2 = -(x)^2
-16 = -16
sqrt -16 = sqrt -x^2 = sqrt (-x * -x)
sqrt -16 = -x, now multiply both sides by negative 1 again
-sqrt 16 = x
-4 = x
sqrt -16 = -4

P.j.S

sqrt -16 = x
x = 4 = sqrt 16
-x = -4
x^2 = 16
-x^2 = -16
-16 = -16
sqrt -16 = sqrt -x^2
sqrt -16 = sqrt (-x * -x)
sqrt -16 = -x = -4
sqrt -16 = -4 = x
x = -4
sqrt -16 = -4

The line -sqrt 16 = x in the original post is proved true.
sqrt -16 = -x times -1 is the same as -sqrt 16 = x or x = -4.

P.j.S
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PJS
post Jul 08, 2010, 10:06 AM
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QUOTE(PJS @ Jul 08, 2010, 09:21 AM) *

QUOTE(PJS @ Jul 07, 2010, 10:44 AM) *

sqrt -16 = x
-sqrt 16 = - x
-4 = -x
-(4)^2 = -(x)^2
-16 = -16
sqrt -16 = sqrt -x^2 = sqrt (-x * -x)
sqrt -16 = -x, now multiply both sides by negative 1 again
-sqrt 16 = x
-4 = x
sqrt -16 = -4

P.j.S

sqrt -16 = x
x = 4 = sqrt 16
-x = -4
x^2 = 16
-x^2 = -16
-16 = -16
sqrt -16 = sqrt -x^2
sqrt -16 = sqrt (-x * -x)
sqrt -16 = -x = -4
sqrt -16 = -4 = x
x = -4
sqrt -16 = -4

The line -sqrt 16 = x in the original post is proved true.
sqrt -16 = -x times -1 is the same as -sqrt 16 = x or x = -4.

P.j.S

x = -4
-x = -4
-x^2 = 16
sqrt (x = -4) * (-x = -4) = sqrt 16
-4 = -4

x^2 = -16
sqrt (x = -4) * (x = -4) = sqrt -16
x = -4 = sqrt -16 See.

sqrt (x = -4) * (-x = -4) = -4 for x or -x.
sqrt -16 = x = -4

The solution for negative square roots in my opinion.

P.j.S
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Homegamek
post Jul 08, 2010, 10:31 AM
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QUOTE(PJS @ Jul 08, 2010, 05:50 PM) *

QUOTE(Homegamek @ Jul 07, 2010, 11:52 AM) *

QUOTE(PJS @ Jul 07, 2010, 07:44 PM) *

sqrt -16 = x
-sqrt 16 = - x
-4 = -x
-(4)^2 = -(x)^2
-16 = -16
sqrt -16 = sqrt -x^2 = sqrt (-x * -x)
sqrt -16 = -x, now multiply both sides by negative 1 again
-sqrt 16 = x
-4 = x
sqrt -16 = -4

P.j.S


I wonder you discard the complex numbers??? Consider what you wrote in frames of conventional rules of the mathematics, I do not understand what you want to say by your writings. Your transition between lines 7 and 8 contains ERROR: see point J.

A)
Input: Solve[Sqrt[-16] == x]
Output: x= 4 * i , where i is the imaginary number (here and below the sign "*" is used as multiplication sign).

cool.gif - Sqrt[16]= - x <=> (-1)* Sqrt[16]= (-1)* x <=> Sqrt[16]= x
Input: Solve[Sqrt[16] == x]
Output: x=4

C) -4 = -x <=> (-1)*4 =(-1)*x

((-1)*4)^2 =((-1)*x )^2 =>
((-1)^2)*(4^2) =((-1)^2)*(x ^2 ) , then
as (-1)^2=1, we have
4^2=x^2

Input: Solve[x^2=4^2]
Output: x1=-4, x2=4 , so the equation has 2 solutions.

D) -16 = -16 = (-1)* (4^2)=(-1)* ((-4)^2)

E1) Sqrt[-16] = Sqrt[(-x)^2] <==> Sqrt[-16] = Sqrt[x^2]

Input: Solve[Sqrt[-16] == Sqrt[(-x)^2]]
Output: x1=-4*i, x2=4*i , where i is the complex number, so the equation has 2 solutions.


E2)Input: Solve[Sqrt[-16] == Sqrt[-x^2]]
Output: x1=-4, x2=4 , so the equation has 2 solutions.

F) Input: Solve[Sqrt[-16] == -x]
Output: x=-4*i , where i is the complex number

J) If you multiply both sides of Sqrt[-16] = -x by (-1) you will get -Sqrt[-16] = x and not -Sqrt[16] = x as you mention.

Input: Solve[-Sqrt[-16] == x]
Output: x= -4*i

x^3 - 2x^2 = 1 proves that imaginary numbers are in error. See x^2 + 1 = 0. x^2 = -1.
x^3 - 2(-1) = 1
x^3 + 2 = 1
x^3 = 1 - 2
x^3 = -1
x = -1 , x^2 = 1 So x^2 = -1 and 1 at the same time. Another arrangement for negative square roots may be needed.

P.j.S


In fact you use traditional operators 's symbols assigning them new properties. If you want to introduce New Mathematical operators, then please invent new ontology. In my next post I shall expound that issue. Below consider PrintScreen image concerning the equations you provided above (please delete spaces as I cannot post URLs yet):

://lh4. ggpht. com/_T8rxhWAaBf8/TDYYM0mGmAI/AAAAAAAAABc/6QqSD0MPx3A/s800/forPjS. JPG
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Homegamek
post Jul 08, 2010, 10:39 AM
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QUOTE(PJS @ Jul 08, 2010, 06:21 PM) *

sqrt -16 = sqrt (-x * -x)
sqrt -16 = -x = -4


Above I mentioned that you use traditional operators assigning them new properties.
You should define your own operators and should not use the traditional ones. It is very important. Look:

sqrt (-x * -x) is not equal to -x because the definition of Operator SQRT is not such.

In fact you must introduce a new operator e.g. sqrtPJS stating that
1. sqrtPJS[-x]=-sqrtPJS[x] if x>0
2. sqrtPJS[(-x)*(-x)]=-x if x>0
3. etc

QUOTE(PJS @ Jul 08, 2010, 06:21 PM) *

The line -sqrt 16 = x in the original post is proved true.
sqrt -16 = -x times -1 is the same as -sqrt 16 = x or x = -4.


Here again the same problem with ontology:
sqrt (-16) is not equal to - sqrt (16) !

While after defining properly the sqrtPJS operator you can write
sqrtPJS (-16) = - sqrt (16)

Certainly later on you should demonstrate how this NEW OPERATOR shall facilitate CORRECT solution of some Mathematical Problems. If you prove to scientific community that that operator is effective to be used and we can solve many problems without usage of the Imaginary Numbers then many shall use that operator.

Can you show that this NEW OPERATOR shall facilitate CORRECT solution of some Mathematical Problems???
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PJS
post Jul 08, 2010, 02:06 PM
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-x^2 is the same as -x*-x = -x^2. eg. 3*3 = 3^2
sqrt -x^2 = sqrt -x * -x = -x.

sqrt -16 = x
x = 4 = sqrt 16
-x = -4
x^2 = 16
-x^2 = -16
-16 = -16
sqrt -16 = sqrt -x^2
sqrt -16 = sqrt (-x * -x)
sqrt -16 = -x = -4
sqrt -16 = x
sqrt -16 = -4 = x
x = -4
sqrt -16 = -4

here again x = -4 is solved without the -sqrt 16 = x line.

P.j.S
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Homegamek
post Jul 08, 2010, 08:31 PM
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Hope when you be apt to converse via usage of English letters and Common Sense, we shall return to this conversation.
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PJS
post Jul 15, 2010, 09:29 AM
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x = -4
-x = -4

-x^2 = -4x
-x^2 = 16
x^2 = -16
x = sqrt -16
-4 = sqrt -16

The idea is to make x = sqrt -16. Then the sqrt of (x = -4) * (-x = -4) is -4 and then -4 = sqrt -16.

sqrt -x^2 = sqrt -x * -x = -x
sqrt -x^2 = sqrt -x * x = x sqrt "-" multiplying by -1 yields -x sqrt "+" for example sqrt -16 times -1 equals
"-" sqrt +16 = -4. Negative 4 is the square root of 16.

P.j.S
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P.j.S
post Dec 23, 2010, 01:19 PM
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3=3
3*(-3/-3)=3
3*-3=3*-3
-9 = 3*-3
sqrt -9 = 3*-3

x^2 + 9 = ?
x^2 = -9
sqrt x^2 = sqrt -9
x = 3*-3
x/-3 = 3
x = -9
(-9)^2 + 9 = 90
? = 90

P.j.S
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P.j.S
post Jan 05, 2011, 12:49 PM
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QUOTE(P.j.S @ Dec 23, 2010, 01:19 PM) *

3=3
3*(-3/-3)=3
3*-3=3*-3
-9 = 3*-3
sqrt -9 = 3*-3

x^2 + 9 = ?
x^2 = -9
sqrt x^2 = sqrt -9
x = 3*-3
x/-3 = 3
x = -9
(-9)^2 + 9 = 90
? = 90

P.j.S

x^2 + 9 = ?
x^2 = -9 + ?
x^2 = ? = 9
x = sqrt ? = 3
x + 3 = sqrt ?
-3 + 3 = sqrt 0
0 = 0
x = -3
x^2 + 9 = ?
(-3)^2 + 9 = 18
? = 18

? = 0
sqrt -9 = -3
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code buttons
post Jan 05, 2011, 06:06 PM
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You need a girlfriend, mate!
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P.j.S
post Jan 06, 2011, 11:35 AM
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QUOTE(code buttons @ Jan 05, 2011, 06:06 PM) *

You need a girlfriend, mate!

I suppose.

?=0

x^2 + 9 = ?
x^2 + 9 = 0
x^2 = 0 - 9
x^2 = 0 = 9
sqrt x^2 = sqrt 0 = sqrt 9
x = 0 = 3
x = 0 - 3
x + 3 = 0
x = -3

But there it is. x = -3.
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P.j.S
post Jan 06, 2011, 11:55 AM
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QUOTE(P.j.S @ Jan 06, 2011, 11:35 AM) *

QUOTE(code buttons @ Jan 05, 2011, 06:06 PM) *

You need a girlfriend, mate!

I suppose.

?=0

x^2 + 9 = ?
x^2 + 9 = 0
x^2 = 0 - 9
x^2 = 0 = 9
sqrt x^2 = sqrt 0 = sqrt 9
x = 0 = 3
x = 0 - 3
x + 3 = 0
x = -3

But there it is. x = -3.

(x + 3)^2
x^2 + 6x + 9
x = -3
9 - 18 + 9 = 0
9 - 9 = 0

x^2 + 9 = 0
sqrt x^2 = x + 3
x = -3
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P.j.S
post Jan 06, 2011, 12:05 PM
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QUOTE(P.j.S @ Jan 06, 2011, 11:55 AM) *

QUOTE(P.j.S @ Jan 06, 2011, 11:35 AM) *

QUOTE(code buttons @ Jan 05, 2011, 06:06 PM) *

You need a girlfriend, mate!

I suppose.

?=0

x^2 + 9 = ?
x^2 + 9 = 0
x^2 = 0 - 9
x^2 = 0 = 9
sqrt x^2 = sqrt 0 = sqrt 9
x = 0 = 3
x = 0 - 3
x + 3 = 0
x = -3

But there it is. x = -3.

(x + 3)^2
x^2 + 6x + 9
x = -3
9 - 18 + 9 = 0
9 - 9 = 0

x^2 + 9 = 0
sqrt x^2 = x + 3
x = -3

x^2 + 1 = 0
sqrt x^2 = x + 1 = 0
x = -1

(x+1)^2
x^2 + 2x + 1 = 0
x = -1
1 - 2 + 1 = 0
1 - 1 = 0
x = -1

The sqrt -1 is 0. This is because...
x^2 + 1 = 0
x^2 = -1
sqrt x^2 = x+1 = sqrt -1
sqrt -1 = x+1
x = -1
x + 1 = -1 + 1 = 0
sqrt -1 = 0

The square root of -1 is x + 1. With x = -1 then the sqrt -1 = 0. 0 is the square root of any negative number.
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P.j.S
post Jan 06, 2011, 03:18 PM
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QUOTE(P.j.S @ Jan 06, 2011, 12:05 PM) *

QUOTE(P.j.S @ Jan 06, 2011, 11:55 AM) *

QUOTE(P.j.S @ Jan 06, 2011, 11:35 AM) *

QUOTE(code buttons @ Jan 05, 2011, 06:06 PM) *

You need a girlfriend, mate!

I suppose.

?=0

x^2 + 9 = ?
x^2 + 9 = 0
x^2 = 0 - 9
x^2 = 0 = 9
sqrt x^2 = sqrt 0 = sqrt 9
x = 0 = 3
x = 0 - 3
x + 3 = 0
x = -3

But there it is. x = -3.

(x + 3)^2
x^2 + 6x + 9
x = -3
9 - 18 + 9 = 0
9 - 9 = 0

x^2 + 9 = 0
sqrt x^2 = x + 3
x = -3

x^2 + 1 = 0
sqrt x^2 = x + 1 = 0
x = -1

(x+1)^2
x^2 + 2x + 1 = 0
x = -1
1 - 2 + 1 = 0
1 - 1 = 0
x = -1

The sqrt -1 is 0. This is because...
x^2 + 1 = 0
x^2 = -1
sqrt x^2 = x+1 = sqrt -1
sqrt -1 = x+1
x = -1
x + 1 = -1 + 1 = 0
sqrt -1 = 0

The square root of -1 is x + 1. With x = -1 then the sqrt -1 = 0. 0 is the square root of any negative number.

Double Checking:

x^2 + 9 = 0
x^2 = 0 - 9
x^2 = 0 = 9
x = sqrt 0 = 3
x = sqrt 0 - 3
x + 3 = sqrt 0
(x + 3)^2 = 0
x^2 + 6x + 9 = 0
x^2 + 9 = -6x
x = -3
9 + 9 = 18
18 - 18 = 0

The answer x = -3 reflects that the middle term 6x is missing from x^2 + 9 = 0.
There is more to x^2 + 9 = 0 for the term "x".
x^2 + 9 = 0
(-3)^2 + 9 = 18
9 + 9 = 18
18 = -6*-3
18 = -6x
(-3)^2 + 9 = -6x
x^2 + 6x + 9 = 0
sqrt (x^2 + 6x + 9) = sqrt 0
x+3 = sqrt 0
x+3 = 0
x = -3
x^2 + 6x + 9 = 0
(-3)^2 + 6(-3) + 9 = 0
9 - 18 + 9 = 0
9 - 9 = 0
0 = 0

sqrt (x^2 + 6x + 9) = x+3
x+3 is the square root of the complete equation x^2 + 6x + 9 = 0.
x = -3

The high light is x+3 = sqrt 0. Squaring both sides reveals the equation with the missing term 6x.
The notation x^2 = -9 is only part of a larger question. To find the square root of x^2 = -9 one must first add the middle term to the equation. Then x = -3 is easier to see as a complete answer.
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post Jan 06, 2011, 07:22 PM
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QUOTE(P.j.S @ Jan 06, 2011, 06:18 PM) *

QUOTE(P.j.S @ Jan 06, 2011, 12:05 PM) *

QUOTE(P.j.S @ Jan 06, 2011, 11:55 AM) *

QUOTE(P.j.S @ Jan 06, 2011, 11:35 AM) *

QUOTE(code buttons @ Jan 05, 2011, 06:06 PM) *

You need a girlfriend, mate!

I suppose.

?=0

x^2 + 9 = ?
x^2 + 9 = 0
x^2 = 0 - 9
x^2 = 0 = 9
sqrt x^2 = sqrt 0 = sqrt 9
x = 0 = 3
x = 0 - 3
x + 3 = 0
x = -3

But there it is. x = -3.

(x + 3)^2
x^2 + 6x + 9
x = -3
9 - 18 + 9 = 0
9 - 9 = 0

x^2 + 9 = 0
sqrt x^2 = x + 3
x = -3

x^2 + 1 = 0
sqrt x^2 = x + 1 = 0
x = -1

(x+1)^2
x^2 + 2x + 1 = 0
x = -1
1 - 2 + 1 = 0
1 - 1 = 0
x = -1

The sqrt -1 is 0. This is because...
x^2 + 1 = 0
x^2 = -1
sqrt x^2 = x+1 = sqrt -1
sqrt -1 = x+1
x = -1
x + 1 = -1 + 1 = 0
sqrt -1 = 0

The square root of -1 is x + 1. With x = -1 then the sqrt -1 = 0. 0 is the square root of any negative number.

Double Checking:

x^2 + 9 = 0
x^2 = 0 - 9
x^2 = 0 = 9
x = sqrt 0 = 3
x = sqrt 0 - 3
x + 3 = sqrt 0
(x + 3)^2 = 0
x^2 + 6x + 9 = 0
x^2 + 9 = -6x
x = -3
9 + 9 = 18
18 - 18 = 0

The answer x = -3 reflects that the middle term 6x is missing from x^2 + 9 = 0.
There is more to x^2 + 9 = 0 for the term "x".
x^2 + 9 = 0
(-3)^2 + 9 = 18
9 + 9 = 18
18 = -6*-3
18 = -6x
(-3)^2 + 9 = -6x
x^2 + 6x + 9 = 0
sqrt (x^2 + 6x + 9) = sqrt 0
x+3 = sqrt 0
x+3 = 0
x = -3
x^2 + 6x + 9 = 0
(-3)^2 + 6(-3) + 9 = 0
9 - 18 + 9 = 0
9 - 9 = 0
0 = 0

sqrt (x^2 + 6x + 9) = x+3
x+3 is the square root of the complete equation x^2 + 6x + 9 = 0.
x = -3

The high light is x+3 = sqrt 0. Squaring both sides reveals the equation with the missing term 6x.
The notation x^2 = -9 is only part of a larger question. To find the square root of x^2 = -9 one must first add the middle term to the equation. Then x = -3 is easier to see as a complete answer.

really though.... you are kind of a moron pal! Good at math maybe but a complete moron nonetheless!
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Trip like I do
post Jan 07, 2011, 08:05 AM
Post #19


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QUOTE(Trip like I do @ Jan 06, 2011, 10:22 PM) *

QUOTE(P.j.S @ Jan 06, 2011, 06:18 PM) *

QUOTE(P.j.S @ Jan 06, 2011, 12:05 PM) *

QUOTE(P.j.S @ Jan 06, 2011, 11:55 AM) *

QUOTE(P.j.S @ Jan 06, 2011, 11:35 AM) *

QUOTE(code buttons @ Jan 05, 2011, 06:06 PM) *

You need a girlfriend, mate!

I suppose.

?=0

x^2 + 9 = ?
x^2 + 9 = 0
x^2 = 0 - 9
x^2 = 0 = 9
sqrt x^2 = sqrt 0 = sqrt 9
x = 0 = 3
x = 0 - 3
x + 3 = 0
x = -3

But there it is. x = -3.

(x + 3)^2
x^2 + 6x + 9
x = -3
9 - 18 + 9 = 0
9 - 9 = 0

x^2 + 9 = 0
sqrt x^2 = x + 3
x = -3

x^2 + 1 = 0
sqrt x^2 = x + 1 = 0
x = -1

(x+1)^2
x^2 + 2x + 1 = 0
x = -1
1 - 2 + 1 = 0
1 - 1 = 0
x = -1

The sqrt -1 is 0. This is because...
x^2 + 1 = 0
x^2 = -1
sqrt x^2 = x+1 = sqrt -1
sqrt -1 = x+1
x = -1
x + 1 = -1 + 1 = 0
sqrt -1 = 0

The square root of -1 is x + 1. With x = -1 then the sqrt -1 = 0. 0 is the square root of any negative number.

Double Checking:

x^2 + 9 = 0
x^2 = 0 - 9
x^2 = 0 = 9
x = sqrt 0 = 3
x = sqrt 0 - 3
x + 3 = sqrt 0
(x + 3)^2 = 0
x^2 + 6x + 9 = 0
x^2 + 9 = -6x
x = -3
9 + 9 = 18
18 - 18 = 0

The answer x = -3 reflects that the middle term 6x is missing from x^2 + 9 = 0.
There is more to x^2 + 9 = 0 for the term "x".
x^2 + 9 = 0
(-3)^2 + 9 = 18
9 + 9 = 18
18 = -6*-3
18 = -6x
(-3)^2 + 9 = -6x
x^2 + 6x + 9 = 0
sqrt (x^2 + 6x + 9) = sqrt 0
x+3 = sqrt 0
x+3 = 0
x = -3
x^2 + 6x + 9 = 0
(-3)^2 + 6(-3) + 9 = 0
9 - 18 + 9 = 0
9 - 9 = 0
0 = 0

sqrt (x^2 + 6x + 9) = x+3
x+3 is the square root of the complete equation x^2 + 6x + 9 = 0.
x = -3

The high light is x+3 = sqrt 0. Squaring both sides reveals the equation with the missing term 6x.
The notation x^2 = -9 is only part of a larger question. To find the square root of x^2 = -9 one must first add the middle term to the equation. Then x = -3 is easier to see as a complete answer.

really though.... you are kind of a moron pal! Good at math maybe but a complete moron nonetheless!

I bet you must either be really really good at doing income tax returns or you are really really bad at doing them!
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P.j.S
post Jan 08, 2011, 06:31 AM
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x^2 + 1 = 0
x^2 = -1
- x^2 = 1
sqrt -x^2 = sqrt 1
sqrt (-x)(-x) = sqrt 1
-x = 1
x = -1
--1 = 1
1 = 1
1 - 1 = 0

x^2 + 1 = 0
(-1)^2 + 1 = 2
(-1)^2 + 1 = -2(-1)
x^2 + 1 = -2x
x^2 + 2x + 1 = 0 Now we have the true equation. Taking the square root...
x + 1 = sqrt 0
x + 1 = 0
x = -1

"Trip" I see in your picture that you probably trip over your lip.
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post Jan 08, 2011, 07:20 AM
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QUOTE(PJS @ Jul 07, 2010, 10:44 AM) *

sqrt -16 = x
-sqrt 16 = - x
-4 = -x
-(4)^2 = -(x)^2
-16 = -16
sqrt -16 = sqrt -x^2 = sqrt (-x * -x)
sqrt -16 = -x, now multiply both sides by negative 1 again
-sqrt 16 = x
-4 = x
sqrt -16 = -4

P.j.S

sqrt -16 = x
-16 = x^2
x^2 + 16 = 0
x = -4 from above

(-4)^2 + 16 = 32
(-4)^2 + 16 = -8(-4)
x^2 + 16 = -8x
x^2 + 8x + 16 = 0
sqrt (x^2 + 8x + 16) = sqrt 0
x + 4 = 0
x = -4

P.j.S
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PJS
post Jan 08, 2011, 08:27 AM
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It appears that x^2 + 1 = 0 is a mathematical error or is incomplete or both.
The exercise shows that x = -1 yet when applied to x^2 the result is 2 not 0 in the equality.
Finishing the equation with a result of 2 shows that it is more correctly written as x^2 + 2x + 1 = 0
The square root of which is x+1.

I hope that you have found this thread to be interesting.

PJS
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post Jan 08, 2011, 01:39 PM
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QUOTE(P.j.S @ Jan 08, 2011, 09:31 AM) *

x^2 + 1 = 0
x^2 = -1
- x^2 = 1
sqrt -x^2 = sqrt 1
sqrt (-x)(-x) = sqrt 1
-x = 1
x = -1
--1 = 1
1 = 1
1 - 1 = 0

x^2 + 1 = 0
(-1)^2 + 1 = 2
(-1)^2 + 1 = -2(-1)
x^2 + 1 = -2x
x^2 + 2x + 1 = 0 Now we have the true equation. Taking the square root...
x + 1 = sqrt 0
x + 1 = 0
x = -1

"Trip" I see in your picture that you probably trip over your lip.


haha.... yeah, quite often
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PJS
post May 15, 2011, 11:53 AM
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x^2 + 1 = 0

x^2 = -1
-x^2 = 1
sqrt -x^2 = sqrt 1
sqrt -x^2 = 1
-sqrt -x^2 = -1 = x^2
-sqrt -x^2 = x^2 therefore
-sqrt - cancels and leaves x^2 = x^2 = -1 Wow!

x^2 + 16 = 0
x^2 = -16
sqrt x^2 = sqrt -16
- sqrt x^2 = -sqrt -16
-sqrt x^2 = 16
sqrt -sqrt x^2 = sqrt 16
sqrt -x = 4
-sqrt -x = -4
x = -4 Wow!

x^2 + 16 = 0
(-4)^2 + 16 = 32
16 + 16 = -8(-4)
x^2 + 16 = -8(x)
x^2 + 8x + 16 = 0
sqrt (x^2 + 8x + 16) = sqrt 0
(x + 4) = 0
x = -4 Wow!
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P JayS
post Apr 24, 2013, 08:52 PM
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Two examples of square roots in my opinion...

sqrt 16 = |4| so (+4)^2 = (-4)^2 = 16

sqrt -16 = -4 so -4 * 4 = -16

so in general terms ...

sqrt -(x^2) = -x so -x * x = -(x^2)

...looking for the solution to sqrt -16 = +4
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P JayS
post Apr 24, 2013, 11:17 PM
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x^2 + 16 = 0
x^2 = -16
-(x^2) = 16
sqrt -(x^2) = 4 or -4

Therefore my assertion that in general terms is correct ...

sqrt -(x^2) = +x so -x * x = -(x^2)

x^2 + 1 = 0
x^2 = -1
-(x^2) = 1
sqrt -(x^2) = sqrt 1 = +1

Therefore the square root -1 = +1.

P.j.S .
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