Solve Simple Square Roots: (-6)^1/2 x (-7)^1/2 = (42)^1/2

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Homework Help Overview

The discussion revolves around the mathematical manipulation of square roots involving negative numbers, specifically the expression \((-6)^{1/2} \times (-7)^{1/2} = (42)^{1/2}\). Participants are exploring the validity of the steps taken in the calculation and the properties of square roots.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning the application of properties of exponents and square roots, particularly when dealing with negative bases. There is a discussion about the validity of combining square roots of negative numbers and the implications of using imaginary numbers.

Discussion Status

Some participants have pointed out potential errors in the original poster's reasoning, particularly regarding the assumption that the property \(x^z y^z = (xy)^z\) holds for negative values. Others have suggested that the right-hand side of the equation should remain unchanged during manipulation.

Contextual Notes

There is an implicit assumption that the properties of square roots apply universally, which is being challenged in the context of negative numbers. The discussion is also constrained by the need to adhere to homework guidelines that may limit the exploration of complex numbers.

frozen7
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((-6)^ 1/2 ) ( (-7)^1/2) = (42)^1/2

((-6)^ 1/2 ) ( (-7)^1/2) = (6^ 1/2 ) ( 7^1/2) ( i ^2)
= -(42)^1/2

Can anyone tell me where I did wrong?
 
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frozen7 said:
((-6)^ 1/2 ) ( (-7)^1/2) = (42)^1/2

((-6)^ 1/2 ) ( (-7)^1/2) = (6^ 1/2 ) ( 7^1/2) ( i ^2)
= -(42)^1/2

Can anyone tell me where I did wrong?
((-6)^ 1/2 ) ( (-7)^1/2) = (42)^1/2
this step
[tex]x^zy^z=(xy)^z[/tex]
is not true in general only when x,y>0
 
((-6)(-7))^(1/2)=(42)^(1/2)
(42)^(1/2)=(42)^(1/2)
should have left the right hand side alone
 
Thanks a lot..
 

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