Just draw a complex plane. 1 is real, then sqrt(x) and sqrt(-x) are complex conjugate. Re(sqrt(x) s/b 1/2, but x = -x that means Re(x) = 0 and sqrt(x) s/b 1/2, so Im(x) = 1/2 => answer is +-i/2
Pretty much the same as for Real numbers. Square root of X is the number that when multiplied by itself results in X. In both cases, unless otherwise specified, we mean the "principal square root", i.e., the root with the non-negative Real component.
Yes, there can be a square root of a negative number if you consider Complex numbers, which is done all the time in science and engineering to solve real world problems.
@@RealMesaMike No. There can not. The square root is a function defined only for nonnegative real numbers, and for a reason. Look at this: 1=sqrt(1)=sqrt((-1)*(-1)) =sqrt(-1)*sqrt(-1)=i*i=-1 Which is nonsense. If we were to define sqrt(-1)=i, the law sqrt(a*b)=sqrt(a)*sqrt(b) would no longer hold. The definition is i^2=-1. Not i=sqrt(-1).
@@andrewd7699 I know. Doesn't matter. Still wrong. I'm so tired of people distributing this false claim. Even scientists do that occasionally when speaking to public in spite of them knowing better. And i'm even more tired of people rather believing sources or higher rank people they don't understand than thinking for themselves. Schools ask you to study 'knowledge' for hours, and they call that 'learning', and they evaluate you by how good you can repeat that knowledge, not by how good you can solve problems. I gave you an explanation on why the square root is defined only for positive real numbers. This explanation is more valuable than any source stating the opposite. If for whatever reason you still rather wanted to contradict me than being open to it, then you need to attack my logic and not come with an argument by authority. In case you're still insisting: They're talking about the principal square root here, not the actual square root. For the principal square root, the exact issue i pointed out exists: you cannot swap the order with multiplication anymore. They even say this in the section, apparently you didn't even read the article. The square root is only defined for nonnegative real numbers.
Just draw a complex plane. 1 is real, then sqrt(x) and sqrt(-x) are complex conjugate. Re(sqrt(x) s/b 1/2, but x = -x that means Re(x) = 0 and sqrt(x) s/b 1/2, so Im(x) = 1/2 => answer is +-i/2
sqrt(x) + sqrt(-x) = 1
sqrt(x) + sqrt(x)i = 1
y = sqrt(x)
y + iy = 1
y(1 + i) = 1
y = 1 / (1 + i)
sqrt(x) = 1 / (1 + i)
x = (1 / (1 + i))^2 = 1 / (1 + 2i + i^2) = 1 / (2i)
sqrt(x) + sqrt(-x) = 1
x + 2*sqrt(x)*sqrt(-x) -x = 1^2 = 1
2*sqrt(-x^2) = 1
2*sqrt(-1*x^2) = 1
2*sqrt(-1)*sqrt(x^2) = 1
2*i*(+/- x) = 1
x = +/- 1/(2*i)
x = +/- i*0.5
Q.E.D.
-x=1-2sqrtx+x , 4x+2=-1 , x= |+/- | sqrt(-1/4) , 4x^2=-1 , x=i/sqrt4 , x= i/2 , i/(-2) ,
Nice❤
The problem is poorly formulated. One needs to specify what is meant by sqrt(x) for complex x. Is the same branch is used for sqrt(x) and sqrt (-x)?
Pretty much the same as for Real numbers. Square root of X is the number that when multiplied by itself results in X.
In both cases, unless otherwise specified, we mean the "principal square root", i.e., the root with the non-negative Real component.
There cannot be a square root of a negative number, so how can both sqrt(x) and sqrt(-x) be possible, unless x = 0?
Yes, there can be a square root of a negative number if you consider Complex numbers, which is done all the time in science and engineering to solve real world problems.
@@RealMesaMike No. There can not. The square root is a function defined only for nonnegative real numbers, and for a reason. Look at this:
1=sqrt(1)=sqrt((-1)*(-1))
=sqrt(-1)*sqrt(-1)=i*i=-1
Which is nonsense.
If we were to define sqrt(-1)=i, the law sqrt(a*b)=sqrt(a)*sqrt(b) would no longer hold.
The definition is i^2=-1.
Not i=sqrt(-1).
@@lindor941 en.wikipedia.org/wiki/Square_root#Square_roots_of_negative_and_complex_numbers
@@andrewd7699 I know. Doesn't matter. Still wrong. I'm so tired of people distributing this false claim. Even scientists do that occasionally when speaking to public in spite of them knowing better.
And i'm even more tired of people rather believing sources or higher rank people they don't understand than thinking for themselves. Schools ask you to study 'knowledge' for hours, and they call that 'learning', and they evaluate you by how good you can repeat that knowledge, not by how good you can solve problems.
I gave you an explanation on why the square root is defined only for positive real numbers. This explanation is more valuable than any source stating the opposite. If for whatever reason you still rather wanted to contradict me than being open to it, then you need to attack my logic and not come with an argument by authority.
In case you're still insisting: They're talking about the principal square root here, not the actual square root. For the principal square root, the exact issue i pointed out exists: you cannot swap the order with multiplication anymore. They even say this in the section, apparently you didn't even read the article. The square root is only defined for nonnegative real numbers.
soluzione corretta di un problema irrisolvibile
your solution is incorrect
I feel same when I saw him unbracket the quadratic equ😅
The solution was fine and the answer is correct.