A Radical Equation With Complex Numbers

🤩 Hello everyone, I'm very excited to bring you a new channel (SyberMath Shorts)
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #RadicalEquations #ComplexNumbers
via @YouTube @Apple @Desmos @NotabilityApp @googledocs @canva

PLAYLISTS 🎵 :

[MathOrient]

There is a simpler solution:

Raise both sides to the power of 2:

x - x + 2sqrt(-x ^2) = 8- 6i
i * (+-x)= 4 - 3i
+-x = -4i -3 ----> x = +-(-4i - 3) = -+(4i + 3)

[Blaqjaqshellaq]

The solution can also presented as x=5*e^[i*arctan(4/3)]

[thomaslangbein]

The positive square root of i^2 is only i. -i is the negative square root of i^2. It mustn’t be confused with x being a solution of y=(x^2)^0.5. It‘s clear that y can be x and -x. Two different things!

[demenion]

you could've just taken out sqrt(i²)=±i in the method presented first. then you would have sqrt(x)=(3-i)/(1±i) which after simplifying gives you 1-2i and 2+i. the first solution squares to -3-4i and the second one to 3+4i

[ayuvell]

-6 - 8i by squaring both sides in my head, now off to watch the video
Edit: yea, forgot to divide by 2, so
-3 - 4i
And I also missed the opposite (3+4i), so thanks for explaining the problem

[thomasbassil]

Just rewrite sqrt(-x) as i × sqrt(x) then factor by sqrt(x) and divide both sides by (1+ i) then square both sidesand reduce

[m]

No matter how many time you say you’re starting with the second method, you’re still starting with the first method. :)

[scottleung]

I got the negative solution, but didn't think of its positive counterpart.

[tetsuyaikeda]

hello why the 1st method is presented at last part?

[alextang]

Can u use polar form to solve this question? I think we can. 😋😋😋😋😋😋

[andrejivonin]

in the first method sqr(-x)=+/- i sqr(x)

[kushaldey]

Wait a second, if you start with 2nd method, doesn't it becomes the first method?