Sweden | You need to know this trick! | A Nice Algebra Problem

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[@grrgrrgrr0202]

This can be done so much faster. Just square immediately:
x + 2✓x✓-x -x = 16
✓x✓-x = 8
Squaring again gives
-x^2 = 64
Which gives
x = +- 8i

[@epsi]

Or just
√(-x²)=8
x = 8/i
x = -8i
Addition is commutative, so if the sum of √(-8i) and √(+8i) is 4, then the sum of √(+8i) and √(-8i) must also be 4, meaning
x = ±8i

[@UncleMarkKnowsHisMaths]

Its 1(rootx) + i(rootx)=4
(1+i)(rootx)=4
Rootx =4/(1+i)
x=[4/(1+i)]^2
x=16/2i
x=8/i??

[@epsi]

Don't forget 8/i = 8/√(-1).
The denominator needs to be rationalized according to some teachers, and that turns out to be helpful here.
That can be done by multiplying the entire fraction by i/i=1, resulting in x=8i/(-1), or x=-8i:
x = 8/i
x = (8/i)(i/i)
x = (8•i)/(i•i)
x = 8i/(i²)
x = 8i/(-1)
x = -8i
And if you're wondering why the final result in the video says x=±8i, keep in mind that if x=-8i, then -x=8i.
Addition is commutative, so you can actually swap x with -x, and it'll still work:
√(-8i) + √(8i) = 4 - x=-8i
√(8i) + √(-8i) = 4 - x=+8i
The video actually provides the math for it, but logic works just as well sometimes.
And if you're wondering about the actual values of √(8i) and √(-8i), look into the polar and exponential forms of complex numbers.
You should find
√(8i) = 2+2i
√(-8i) = 2-2i

[@prollysine]

Vx + iVx =4 , let Vx=u , u+iu=4 , u=4/(1+i) , Vx = 4/(1+i) , x=16/(1+i)^2 , x=8/i , *(i/i) --> x= - 8i ,
test , V(-8i) + V8i = 2-2i + 2+2i , 2-2i + 2+2i = 2+2 , 2+2=4 , same , OK ,