Question

for all real x , why is........

Answer 1

\[\sqrt{x}=modx\]

Answer 2

?

Answer 3

i mean and not +/- x

Answer 4

sorry that was x^2

Answer 5

under the root

Answer 6

it becomes more clear as to why this is when u begin dealing with complex numbers, but the square root implies that only one value is being considered, so taking absolute values does not affect the graph of the square root of x

Answer 7

i know a bit about complex numbers ... can you please tell me the basic logic behind it??

Answer 8

idk what type of answer you're looking for, the best I can think of would be because if you take the square root of say, 4, and a particular solution requires x>0, then we would use that to indicate that negative values can't be used. Also note that: \[|a| = a^2\]
whatever number I sub for "a", it will be positive.

Answer 9

after thinking about this problem i have finally made the following conclusion....

from the complex numbers we have....

\[\sqrt{ab}\neq \sqrt{a}*\sqrt{b} foralla,b <0\]

Answer 10

hence we have....\[\sqrt{a ^{2}}=\sqrt{\left| a \right|*\left| a \right|}\]
since IaI>0
therefore.....\[\sqrt{\left| a \right|*\left| a \right|}=\sqrt{\left| a \right|}*\sqrt{\left| a \right|}=(\sqrt{\left| a \right|})^{2}\]

Answer 11

=IaI

but \[\sqrt{a ^{2}}=\sqrt{-\left| a \right|*-\left| a \right|}\neq \sqrt{-\left| a \right|}*\sqrt{-\left| a \right|}\neq(\sqrt{-\left| a \right|})^{2}\neq-\left| a \right|\]

Answer 12

here note that i have used the first property discussed before.....


here in particular ......\[\sqrt{a ^{2}}\neq (\sqrt{a})^{2}\]

Answer 13

if a <0 and therefore \[\sqrt{a ^{2}}=\left| a \right|\]