Solving A Nice Radical Equation

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[tonyhasler]

It is all about the principal square roots of i and -i.
(sqrt(2)/2)(1 +i) and -(sqrt(2)/2)(i+i) both give us i when squared. Similarly, (sqrt(2)/2)(1-i) and -(sqrt(2)/2)(1-i) both give -i when squared. But the principal square roots are the ones with the largest real components, i.e. (sqrt(2)/2)(1+i) and (sqrt(2)/2)(1-i). These do, in fact, add to sqrt(2).

[jir_UwU]

the ± symbol written with the minus on top 😨

[GillesF]

I'm not sure if a negative number can be written under a square root symbol. I would prefer √x + i√x = √2 rather than √x + √(-x) = √2 And from √x + i√x = √2 there is only one root: x = -i

√x + i√x = √2

√x·(1 + i) = √2

1 + i = √2/√x

(1 + i)² = 2/x

1 - 1 + 2i = 2/x

2i = 2/x

2ix = 2

ix = 1

x = 1/i

note: 1/i = (1/i)·(i/i) = (1·i)/(i·i) = i/i² = i/-1 = -i

x = -i

---

/// final result:

■ x = -i

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/// to check √x + i√x = √2 with x = -i:

√(-i) + i√(-i) = 1/√2 - i/√2 + i/√2 - (-1/√2) = 2/√2 = √2 <-- ok

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/// to check √x + i√x = √2 with x = i:

√i + i√i = 1/√2 + i/√2 - 1/√2 + i/√2 = 2i/√2 = i√2 <-- ko !!!

🙂

[bobkurland]

I used substitution x = r*e^(iu) and for -x = e^(ipi)*r*e^(iu) , set that which multiplied i on the left side = 0, solved for r and u to get the same answer x=i

[DrBenway]

I'm sorry but I don't see how x = i ( or x = -i ) can be a solution.
When I substitute x = i into sqrt(x) + sqrt(-x) I get the imaginary number sqrt(2)*i , not the real number sqrt(2).
Anybody else...?

[rainerzufall]

6:17 ERROR ❌ √(i²) = +i (never -i, that is called -√(i²)). But x² = i² leeds to two solutions x = ±i ...

[逸園-無毒果園]

Do you check it?
When x=i or x=-i, we have √i is √2/2 + √2/2i and √-i is -√2/2 + √2/2i.
So √i + √-i =√2i.

[Skank_and_Gutterboy]

The solution I got is: x=(±i)^n where n is {all ±odd integers}.