Negative Square Roots by PJS 
Negative Square Roots by PJS 
PJS 
Jul 07, 2010, 01:44 PM
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#1

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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 
Homegamek 
Jul 07, 2010, 02:52 PM
Post
#2

Aspiring Group: Basic Member Posts: 99 Joined: Jun 26, 2010 Member No.: 32873 
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).  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 
PJS 
Jul 08, 2010, 11:50 AM
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#3

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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).  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 
PJS 
Jul 08, 2010, 12:21 PM
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#4

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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 
PJS 
Jul 08, 2010, 01:06 PM
Post
#5

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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 
Homegamek 
Jul 08, 2010, 01:31 PM
Post
#6

Aspiring Group: Basic Member Posts: 99 Joined: Jun 26, 2010 Member No.: 32873 
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).  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 
Homegamek 
Jul 08, 2010, 01:39 PM
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#7

Aspiring Group: Basic Member Posts: 99 Joined: Jun 26, 2010 Member No.: 32873 
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 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??? 
PJS 
Jul 08, 2010, 05:06 PM
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#8

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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 
Homegamek 
Jul 08, 2010, 11:31 PM
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#9

Aspiring Group: Basic Member Posts: 99 Joined: Jun 26, 2010 Member No.: 32873 
Hope when you be apt to converse via usage of English letters and Common Sense, we shall return to this conversation.

PJS 
Jul 15, 2010, 12:29 PM
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#10

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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 
P.j.S 
Dec 23, 2010, 04:19 PM
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#11

Overlord Group: Basic Member Posts: 358 Joined: Jun 12, 2009 Member No.: 32189 
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 
P.j.S 
Jan 05, 2011, 03:49 PM
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#12

Overlord Group: Basic Member Posts: 358 Joined: Jun 12, 2009 Member No.: 32189 
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 
code buttons 
Jan 05, 2011, 09:06 PM
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#13

Supreme God Group: Basic Member Posts: 2453 Joined: Oct 05, 2005 Member No.: 4556 
You need a girlfriend, mate!

P.j.S 
Jan 06, 2011, 02:35 PM
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#14

Overlord Group: Basic Member Posts: 358 Joined: Jun 12, 2009 Member No.: 32189 

P.j.S 
Jan 06, 2011, 02:55 PM
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#15

Overlord Group: Basic Member Posts: 358 Joined: Jun 12, 2009 Member No.: 32189 
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 
P.j.S 
Jan 06, 2011, 03:05 PM
Post
#16

Overlord Group: Basic Member Posts: 358 Joined: Jun 12, 2009 Member No.: 32189 
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. 
P.j.S 
Jan 06, 2011, 06:18 PM
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#17

Overlord Group: Basic Member Posts: 358 Joined: Jun 12, 2009 Member No.: 32189 
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. 
Trip like I do 
Jan 06, 2011, 10:22 PM
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#18

Supreme God Group: Basic Member Posts: 5157 Joined: Aug 11, 2004 From: Earth^2 Member No.: 3202 
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! 
Trip like I do 
Jan 07, 2011, 11:05 AM
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#19

Supreme God Group: Basic Member Posts: 5157 Joined: Aug 11, 2004 From: Earth^2 Member No.: 3202 
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! 
P.j.S 
Jan 08, 2011, 09:31 AM
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#20

Overlord Group: Basic Member Posts: 358 Joined: Jun 12, 2009 Member No.: 32189 
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. 
P.j.S 
Jan 08, 2011, 10:20 AM
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#21

Overlord Group: Basic Member Posts: 358 Joined: Jun 12, 2009 Member No.: 32189 
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 
PJS 
Jan 08, 2011, 11:27 AM
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#22

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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 
Trip like I do 
Jan 08, 2011, 04:39 PM
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#23

Supreme God Group: Basic Member Posts: 5157 Joined: Aug 11, 2004 From: Earth^2 Member No.: 3202 
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 
PJS 
May 15, 2011, 02:53 PM
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#24

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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! 
P JayS 
Apr 24, 2013, 11:52 PM
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#25

DemiGod Group: Basic Member Posts: 588 Joined: Apr 04, 2012 Member No.: 34146 
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 
P JayS 
Apr 25, 2013, 02:17 AM
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#26

DemiGod Group: Basic Member Posts: 588 Joined: Apr 04, 2012 Member No.: 34146 
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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