5:48, Why do you like to write down (A - B) first? I prefer to write (A + B) first because if A & B are perfect squares then the second factor factors further and the final answer ends up in the traditional order more naturally. For example, x^4 - y^4 = (x^2 + y^2)(x^2 - y^2) = (x^2 + y^2)(x + y)(x - y).
Before watching the video. The fourth roots of 1 are: 1, -1, i, -i So the four solutions to our equation are those numbers times the fourth root of -64, respectively. The fourth root of -64 is the square root of its square root. This can be written as sqrt(sqrt(-64)) = sqrt(sqrt(64)*sqrt(-1)) = sqrt(8i). The square root of i is known as sqrt(0.5) + sqrt(0.5)*i. So the square root of 8i is sqrt(8)(sqrt(0.5) + sqrt(0.5)*i). Simplifying gets you sqrt(4) + sqrt(4)*i = 2 + 2i So the four solutions are: 1(2 + 2i) -1(2 + 2i) i(2 + 2i) -i(2 + 2i) Or simplified: 2 + 2i -2 - 2i -2 + 2i 2 - 2i Using a calculator confirms that these are correct, and you expect exactly 4 answers for a x^4 equation. This can also be written as +-2 +-2i After watching the video: It's very interesting how many legitimate ways there are to solve this!
Why not: x^4+64 = (x^2 -8i)(x^2+8i) = 0 First bracket: x^2 = 8i 8i = 8e^(i*pi/2) Square root halves the angle and reduces the magnitude, so in becomes 2*sqrt(2) * (cos(45)+/-i*sin(45)) = 2*sqrt(2) * (sqrt(2)/2 +/-isqrt(2)/2) = 2+/-2i With second bracket we work analogously. And get the same result as in video. (mor imaginary units will appear but they should turn into minus ones)
you could easily substitute x^2 with u and get two easy solutions with the quadratic formula and when you change back to x^2 you get two solutions from each of the first two.
I would do this as x^4 = -64 = 64 e^(k.i.pi) where k = 1,3,5,7. I know i need 4 values of k as there will be 4 roots. So x = 2sqrt(2) * e^(k/4.i.pi) = 2sqrt(2) * [1+i, -1+i, -1-i, 1-i]/sqrt(2) putting the various k options into [ ] brackets. Therefore x = [2+2i, -2+2i, -2-2i, 2-2i]
i^n. 4throot 64 (for all n={1..4}) Just as for squared equations you need +/- for a fourth power you need to take all possibilities if the fourth root of unity.
Since you are going to talk about complex numbers anyway, it's a lot easier to factor a² + b² when noting that a² + b² = a² - (-b²) = a² - (i²b²) = (a + ib)(a - ib) With a fourth power binomial as here, you end up with two sums/differences of squares that can be factored again as differences/sums of squares.
instead of taking the fourth root of both sides couldnt you instead take the square root twice and do plus or minus both times? You should get ±sqrt(±sqrt(-64)) which when worked out would give the same four answers?
When I was in mid-school and started to learn more math, I was always wary of the "cool" tricks. It was always looming "what if" to me. In this case, the past me would've immediately asked: "hey, what if that 64 was 63?". So, with the amount of effort you put into it - I feel like the moment you talk "complex numbers" you could've briefly introduce the polar form and exponents and then talk Euler's formula. Come to think of it, some cool geometry would be there too - and it would actually solve it for any number, not just special ones.
But why doesn't it work to just take the root, like shown at the start of the video? Is it because we have a function of the 4th degree and the first "method" only gives us 2 instead of 4 possible solutions?
It does work, but you have to apply rules that seem arbitrary to produce all 4 answers. This channel is intended for students that are learning _algebra_, and this is the correct approach for this group of people. If you are learning calculus then you can use more advanced techniques to properly calculate all solutions of the nth root and produce the same answers.
Does this work two, or am I off? x^4 + 64 = 0 x^4 + 64 - 64 = 0 - 64 x^4 = -64 sqrt(x^4) = +/-sqrt(-64) x^2 = +/-sqrt(8*-1) x^2 = +/-sqrt(8)*sqrt(-1) x^2 = +/-sqrt(8)*i x^2 = sqrt(8)*i or x^2 = -sqrt(8)*i sqrt(x^2) = +/-sqrt(sqrt[8]*i), or sqrt(x^2) = +/-sqrt(-sqrt[8]*i) x = +/-sqrt(sqrt[8]*i), or x = +/-sqrt(-sqrt[8]*i) x = +/-sqrt(sqrt[8]*i), or x = +/-sqrt(sqrt[8]*i*[-1]) x = +/-sqrt(sqrt[8])*sqrt(i), or x = +/-sqrt(sqrt[8])*sqrt(i)*sqrt(-1) x = +/-sqrt(sqrt[8])*sqrt(i), or x = +/-sqrt(sqrt[8])*sqrt(i)*i x = +/-(8^[1/2])^(1/2)*i^(1/2), or x = +/-(8^[1/2])^(1/2)*i^(1/2)*i^(2/2) x = +/-8^([1/2]*[1/2])*i^(1/2), or x = +/-8^([1/2]*[1/2])*i^(1/2+2/2) x = +/-8^(1/4)*i^(1/2), or x = +/-8^(1/4)*i^(3/2) x = +/-(2^3)^(1/4)*i^(1/2), or x = +/-(2^3)^(1/4)*i^(3/2) x = +/-2^(3*1/4)*i^(1/2), or x = +/-2^(3*1/4)*i^(3/2) x = +/-2^(3/4)*i^(1/2), or x = +/-2^(3/4)*i^(3/2) x = 2^(3/4)*i^(1/2), or x = -2^(3/4)*i^(1/2), or x = 2^(3/4)*i^(3/2), or x = -2^(3/4)*i^(3/2)
Solve x^(2/3)=64: ruclips.net/video/JzfO9dI4aIk/видео.html
*_"Believe in geometry, not 4throoting both sides"_* 🗣🗣🔥🔥🔥🔥
😆
5:48, Why do you like to write down (A - B) first? I prefer to write (A + B) first because if A & B are perfect squares then the second factor factors further and the final answer ends up in the traditional order more naturally. For example, x^4 - y^4 = (x^2 + y^2)(x^2 - y^2) = (x^2 + y^2)(x + y)(x - y).
"Traditional order"?
Before watching the video.
The fourth roots of 1 are:
1, -1, i, -i
So the four solutions to our equation are those numbers times the fourth root of -64, respectively.
The fourth root of -64 is the square root of its square root. This can be written as sqrt(sqrt(-64)) = sqrt(sqrt(64)*sqrt(-1)) = sqrt(8i). The square root of i is known as sqrt(0.5) + sqrt(0.5)*i. So the square root of 8i is sqrt(8)(sqrt(0.5) + sqrt(0.5)*i). Simplifying gets you sqrt(4) + sqrt(4)*i = 2 + 2i
So the four solutions are:
1(2 + 2i)
-1(2 + 2i)
i(2 + 2i)
-i(2 + 2i)
Or simplified:
2 + 2i
-2 - 2i
-2 + 2i
2 - 2i
Using a calculator confirms that these are correct, and you expect exactly 4 answers for a x^4 equation.
This can also be written as +-2 +-2i
After watching the video:
It's very interesting how many legitimate ways there are to solve this!
Or simply:
i = cos(90°) + i sin(90°)
And therefore:
√i = cos(90°/2) + i sin(90°/2)
6:55: This is a perfect opportunity to use the pq formula instead. You get x₁,₂ = 2 ± √(2² − 8) = 2 ± √−4 = 2 ± 2i
Why not:
x^4+64 = (x^2 -8i)(x^2+8i) = 0
First bracket: x^2 = 8i 8i = 8e^(i*pi/2)
Square root halves the angle and reduces the magnitude, so in becomes 2*sqrt(2) * (cos(45)+/-i*sin(45)) = 2*sqrt(2) * (sqrt(2)/2 +/-isqrt(2)/2) = 2+/-2i
With second bracket we work analogously. And get the same result as in video. (mor imaginary units will appear but they should turn into minus ones)
you could easily substitute x^2 with u and get two easy solutions with the quadratic formula and when you change back to x^2 you get two solutions from each of the first two.
64 = -64i^2 substitute giving x^4 - 64i^2
The factorization then becomes a difference of squares
(x^2 -8i) * (x^2 + 8i)
x^4= -64
x^2=+-8i
x=+-sqrt(+-8i)
And since
i = cos(90°) + i sin(90°),
√i = cos(45°) + i sin(45°)
I would do this as x^4 = -64 = 64 e^(k.i.pi) where k = 1,3,5,7. I know i need 4 values of k as there will be 4 roots.
So x = 2sqrt(2) * e^(k/4.i.pi)
= 2sqrt(2) * [1+i, -1+i, -1-i, 1-i]/sqrt(2) putting the various k options into [ ] brackets.
Therefore x = [2+2i, -2+2i, -2-2i, 2-2i]
i^n. 4throot 64 (for all n={1..4})
Just as for squared equations you need +/- for a fourth power you need to take all possibilities if the fourth root of unity.
Input
(-2 - 2 i)^4 + 64 = 0
Result
True
Since you are going to talk about complex numbers anyway, it's a lot easier to factor a² + b² when noting that
a² + b² = a² - (-b²) = a² - (i²b²) = (a + ib)(a - ib)
With a fourth power binomial as here, you end up with two sums/differences of squares that can be factored again as differences/sums of squares.
makes sense
instead of taking the fourth root of both sides couldnt you instead take the square root twice and do plus or minus both times? You should get ±sqrt(±sqrt(-64)) which when worked out would give the same four answers?
Yeah, i think that works
When I was in mid-school and started to learn more math, I was always wary of the "cool" tricks. It was always looming "what if" to me. In this case, the past me would've immediately asked: "hey, what if that 64 was 63?". So, with the amount of effort you put into it - I feel like the moment you talk "complex numbers" you could've briefly introduce the polar form and exponents and then talk Euler's formula. Come to think of it, some cool geometry would be there too - and it would actually solve it for any number, not just special ones.
But why doesn't it work to just take the root, like shown at the start of the video? Is it because we have a function of the 4th degree and the first "method" only gives us 2 instead of 4 possible solutions?
exactly
i^4 = 1 rather than -1.
It does work, but you have to apply rules that seem arbitrary to produce all 4 answers.
This channel is intended for students that are learning _algebra_, and this is the correct approach for this group of people. If you are learning calculus then you can use more advanced techniques to properly calculate all solutions of the nth root and produce the same answers.
64 = -64i^2 substitute giving x^4 - 64i^2
The factorization then becomes a difference of squares
(x^2 -8i) * (x^2 + 8i)
Does the Po-Shen Loh method work with a quartic like this?
Very nice.
(-2-2i)^4+64=0 x=±2±2i It’s in my head.
Good one❤
Does this work two, or am I off?
x^4 + 64 = 0
x^4 + 64 - 64 = 0 - 64
x^4 = -64
sqrt(x^4) = +/-sqrt(-64)
x^2 = +/-sqrt(8*-1)
x^2 = +/-sqrt(8)*sqrt(-1)
x^2 = +/-sqrt(8)*i
x^2 = sqrt(8)*i or x^2 = -sqrt(8)*i
sqrt(x^2) = +/-sqrt(sqrt[8]*i), or sqrt(x^2) = +/-sqrt(-sqrt[8]*i)
x = +/-sqrt(sqrt[8]*i), or x = +/-sqrt(-sqrt[8]*i)
x = +/-sqrt(sqrt[8]*i), or x = +/-sqrt(sqrt[8]*i*[-1])
x = +/-sqrt(sqrt[8])*sqrt(i), or x = +/-sqrt(sqrt[8])*sqrt(i)*sqrt(-1)
x = +/-sqrt(sqrt[8])*sqrt(i), or x = +/-sqrt(sqrt[8])*sqrt(i)*i
x = +/-(8^[1/2])^(1/2)*i^(1/2), or x = +/-(8^[1/2])^(1/2)*i^(1/2)*i^(2/2)
x = +/-8^([1/2]*[1/2])*i^(1/2), or x = +/-8^([1/2]*[1/2])*i^(1/2+2/2)
x = +/-8^(1/4)*i^(1/2), or x = +/-8^(1/4)*i^(3/2)
x = +/-(2^3)^(1/4)*i^(1/2), or x = +/-(2^3)^(1/4)*i^(3/2)
x = +/-2^(3*1/4)*i^(1/2), or x = +/-2^(3*1/4)*i^(3/2)
x = +/-2^(3/4)*i^(1/2), or x = +/-2^(3/4)*i^(3/2)
x = 2^(3/4)*i^(1/2), or x = -2^(3/4)*i^(1/2), or x = 2^(3/4)*i^(3/2), or x = -2^(3/4)*i^(3/2)
Looks good
Amazing video! this is love from the 8th graders haha
I'm usually a math wizz, this was wild to witness lol
This video must to be called,
How to demostrate the sqrt(i) to non caclulus studients!!!
Why ±√-64 is not amongst the answers?
Because
1) it is a 4th power
2) calculating with ⁴√-64 to directly ±2 ±2i is sloppy maths ;)
x⁴+64=0
(x²)²+8²=0
(x²)²+16x²+8²-16x²=0
(x²+8)²-(4x)²=0
(x²+4x+8)(x²-4x+8)=0
x²+4x+8=0
x²+4x+4=-4
(x+2)²=-4
|x+2|=2i
x+2=±2i
x=-2±2i ❤❤
x²-4x+8=0
x²-4x+4=-4
(x-2)²=-4
|x-2|=2i
x-2=±2i
x=2±2i ❤❤
x^4 - 64i^2 = 0 is easier - difference of squares requires no geometry
I have never seen the geometric approuch to the difference of squares before, thanks man
Same here. My mind is blown.
Initially, I thought "why don't you just do the ±sqrt twice or just the fourth root but with the 1, i, -1, -i attatched too"
I think that's a valid destination of the same answer
noice! 2nd comment btw
It has no solutions, Imaginary numbers don't exist.
They kinda do exist.