What is the difference between √-4^2 and √(-4)^2

[Guest]

So [tex]\sqrt{-4^2}[/tex] simplifies to an imaginary number and [tex]\sqrt{(-4)^2}[/tex] simplifies to a real number.

How and why does this happen?

[Guest]

[tex]-1 = i^2[/tex]
[tex]\sqrt{-4^2} = \sqrt{-1 \cdot 4^2} = \sqrt{i^4 \cdot 2^2} = \sqrt{(4i)^2} = 4i[/tex]

[Guest]

[tex]\sqrt{(-4)^2} = \sqrt{(4)^2} = 4[/tex]

Is it clear?

[Guest]

(-4)^2= (-4)(-4)= 16

-4^2= -(4)(4)= -16.

It is a question of "precedence of operators". Without the parentheses, you first square 4, getting 16 then take the negative of 16 to get -16.

With the paretheses you do what is in the parentheses, to get -4, then square that. The square of -4 is 16.

[Guest]

(-4)^2= (-4)(-4)= 16

-4^2= -(4)(4)= -16.

It is a question of "precedence of operators". Without the parentheses, you first square 4, getting 16 then take the negative of 16 to get -16.

With the paretheses you do what is in the parentheses, to get -4, then square that. The square of -4 is 16.

This makes sense. Thanks!