Question

Again confusion..

Answer 1

\[\large{x^2=1, \textbf{then x = ? }}\]
\[large{x=\sqrt{1}, \textbf{then x = ?}}\]

Answer 2

*\[\large{x=\sqrt{1}, \textbf{then x = ?}}\]

Answer 3

in the first case we have : \(x= \pm 1\)

Answer 4

but in the second case we have : x = 1...

Correct?

Answer 5

no confusion.
x^2=1 is quadratic, it must have 2 roots, 1 and -1
x=root1 is linear, it must have only one root = 1

Answer 6

Ok so, now let us take the second equation :

\[\large{x=\sqrt{1}}\]
\[\large{x^2=(\sqrt{1})^2}\]
\[\large{x^2=1}\]

Answer 7

correct?

Answer 8

when you are squaring, u are attaching one more root to x, (x=-1)

Answer 9

@mathslover i am praying to 3 gods for you :P

Answer 10

yep what @hartnn said is right

Answer 11

thats the fallacy here

Answer 12

I didn't get you exactly @hartnn ?

Answer 13

@mathslover when you squared in 2nd step you also attach (-1)^2 = x^2

Answer 14

\[x^2=1\]\[x=\pm\sqrt 1\]

Answer 15

Oh yes right @UnkleRhaukus

@AravindG sorry again .... any example?

Answer 16

@unkle is right

Answer 17

Why did you include pm there @UnkleRhaukus ?

Answer 18

It's just a matter of notation, √1 refers to the principal square root of 1, which is 1. But in actuality there are two solutions to the equation x=√1, -1 and 1.

Answer 19

Confused..

Answer 20

@dape x=\[\sqrt 1\] has only one solution =1

Answer 21

x^2=1 has two solutions which are -1 and +1

Answer 22

@mathslover just use the theory of equations

Answer 23

i agree with arvind

Answer 24

\[x^2=1\]
\[x_{1,2}=\sqrt 1, -\sqrt1\]\[\qquad=\pm\sqrt 1\]

Answer 25

as you can see a x^2 polynomial will have two roots

Answer 26

let x=a
now when your are squaring, u do
x*x=a*a--->x^2=a^2
but here u also attach here x=-a
so
x*x=(-a)*(-a)
which also is the solution of x^2=a^2

Answer 27

x^2 only has an inverse if we restrict the domain

Answer 28

It's just notation, in actuality the nth root of a number x is \(defined\) to be a number y, which when raised to the nth power is x, or
\[ y^n=x \]
According to this definition, which is the only one i've ever seen. Both -1 and 1 is a 2nd root (square root) of 1.
We usually just take the symbol √ to mean the principal square root (the positive root).

Answer 29

if we ignore it as a "fucntion" then sqrt(x) is