Can we NOT solve x^4+64=0 like this?

[@bprpmathbasics]

Solve x^(2/3)=64: kzbin.info/www/bejne/gKvJgGyafpmUf80

[@Taric25]

@@bprpmathbasics x^⅔=64=x^⅓²=8² ⇔ x⅓²-8²=0=(x^⅓+8)(x^⅓-8) ⇔ x^⅓=±8 ⇔ (±8)³=x=±512.

[@PlzDoNotTheFade]

*_"Believe in geometry, not 4throoting both sides"_* 🗣🗣🔥🔥🔥🔥

[@bprpmathbasics]

😆

[@andresrebolledobanquet1924]

F(x) x^4 = 4x^3 F'(x)4x^3 = 12x^2 F"(x) 12x^2 = 24x 24x + 64 24x = - 64 x = - 64 / 24 x = - 2.66667 - 2.66667*4*3*2 + 64 = 0 - 64 + 64 = 0

[@penqueent13]

​@@andresrebolledobanquet1924what??

[@LordBlee]

Just a couple of comments. I've been an engineer for ~35 years and this is the first time I've seen a geometric explanation of completing the square. This is truly the best explanation I've ever seen. Next, your explanation is nice and detailed and easy to understand. This certainly beats my profs who would in physics would just say "the magic happens here" or mathematicians would say, "this step is left as an exercise to for the student." And in both cases, we'd have about 10 more pages of work...... I really enjoyed this. Thank you for sharing.

[@WillCamx]

I have an Honours degree in pure and applied maths and a Master's from 36 years ago and I was never shown the geometrical explanation of completing the square. Never too old to learn something new.

[@bprpmathbasics]

I am happy to hear this. Thank you so much!

[@LordBlee]

@@WillCamx agreed. I have always loved the old axiom, a day without learning is a day wasted. Lucky for me, 8 always have lots of opportunities.

[@rbarnes4076]

Likewise for me.. this is the first time I've seen this explanation. I've been an engineer for 40 years now! Love learning new things!

[@mhwse]

We do that in Algorithmic Math, where ever it is possible. But still this is a pretty nice example, and the declaration and way of presentation are very clean.

[@rogierownage]

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!

[@Nikioko]

Or simply: i = cos(90°) + i sin(90°) And therefore: √i = cos(90°/2) + i sin(90°/2)

[@Taric25]

It's actually not ±2±2i, because that would only give 2+2i or -2-2i. You would need to state the first one as ±2 and then plus or minus and minus or plus 2i, which is a waste, as you could just say -2±2i or 2±2i.

[@Nikioko]

@@Taric25 That would be ± (2 + 2i).

[@Taric25]

@@Nikioko Also incorrect, since that would exclude -2+2i and 2-2i

[@Nikioko]

@@Taric25 That's exactly what I said. If you only want 2 + 2i and −2 − 2i (like you wrote), ± (2 + 2i) is the way to write it. But ±2 ± 2i is all four solutions.

[@boriscat1999]

I love the geometry example, a very intuitive way to look at this problem

[@Nikioko]

6:55: This is a perfect opportunity to use the pq formula instead. You get x₁,₂ = 2 ± √(2² − 8) = 2 ± √−4 = 2 ± 2i

[@bprpmathbasics]

Ah, I should have done that.

[@brianwade4179]

I solved it entirely using factoring via difference of squares. I started with x^4 - (-64) = 0 and factored that to (x^2+8i)(x^2-8i)= 0. I then wrote x^2+8i as x^2-(-8i) and factored that to (x-(-2+2i))(x+(-2+2i)). This gave me roots -2+2i and 2-2i. I then factored x^2-8i into (x-(2+2i))(x+(2+2i)). This gave me roots 2+2i and -2-2i. Very helpful in this was knowing sqrt(i) = (1+i)/sqrt(2) and sqrt(-i) = (1-i)/sqrt(2).

[@ironicstupidity7102]

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?

[@rogierownage]

Yeah, i think that works

[@Taric25]

That wouldn't work, since it would still only give two solutions. You would need to indicate the second plus or minus and minus or plus, which is a waste, since you could just write √±√64 or -√±√-64.

[@safestate8750]

I have never seen the geometric approuch to the difference of squares before, thanks man

[@pluto9000]

Same here. My mind is blown.

[@Masteroogway-e4i]

Welp... they teach these "algebraic identities "from 6th to 10th grade...

[@eisagdix6692]

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?

[@breqbs]

exactly

[@ronaldking1054]

i^4 = 1 rather than -1.

[@jmr5125]

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.

[@phoonjzc]

Can he come up with a calculus answer​@@jmr5125

[@Goodwalker720]

You can’t get an even root of a negative number, as multiplying two negatives makes a positive product.

[@ukdavepianoman]

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]

[@MrConverse]

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).

[@tfg601]

"Traditional order"?

[@forcelifeforce]

(A + B) first is *not* a traditional order. It is a preferred order for *you.* Do not try to impose your order on everyone else.

[@RobertWyesham]

Other commenters have linked this example to Sophie Germain. Sophie Germaine’s Identity is x^4 + 4y^4 = (x^2 + 2xy + 2y^2)*(x^2 - 2xy + 2y^2). This can be proved by factoring the difference of squares, as follows: x^4 + 4y^4 = x^4 + 4x^2y^2 + 4y^4 - 4x^2y^2 = (x^2 + 2y^2)^2 - (2xy)^2 = (x^2 + 2y^2 + 2xy)*(x^2 + 2y^2 - 2xy) = (x^2 + 2xy + 2y^2)*(x^2 - 2xy + 2y^2) x^4 + 64.= x^4 + 4*16 = x^4 + 4*2^4 = (x^2 + 2*x*2 + 2*2^2)*(x^2 - 2*x*2 + 2*2^2) = (x^2 + 4x + 8)(x^2 - 4x + 8), just as @blackpenredpen has.

[@shannonmcdonald7584]

This is great. I love this geometry. U make it look so clear

[@ssalmero]

Using polar coordinates and extracting the fourth root you get: x=2sqrt(2).e^pi(2n+1)/4, n=0,1,2,3

[@rpocc]

Hm. My first though when looking at the equation is that it’s roots are points at complex plane, sitting exactly at 45°, 135°, -45° and -135° of a circle having radius of sqrt(8). Since sin(45°) = 2^(-0.5) and sqrt(8) = 2^(1.5) I would guess that answers are 2+2i; 2-2i; 2i-2; -2-2i So, you just raise modulus to the 4th degree and multiply the argument by 4 to check the solution: all of those should eventually come to real -64. In opposite, you can take points at real -64 and 1, then calculate four possible angles less than 720 between these to points, divide each by 4 and take a 4th degree root from 64 to determine these four points just in your head.

[@thomasdegroat6039]

I was never taught to compete the square so this is blowing my mind

[@VividhBista]

4:40 what did he say 😭

[@Astrobrant2]

Wow! I've never seen anyone explain completing the square using actual squares and rectangles. Fascinating! Thanks.

[@trueriver1950]

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.

[@EarlyLatte]

I love the cuts that you make xD

[@Simeulf]

The meaning of completing the square makes so much more sense now!! 😮 It is literally completing the square!! Like literally

[@bprpmathbasics]

Indeed! 😆

[@bengt-goranpersson5125]

6:18 - That transition was so smooth!

[@bprpmathbasics]

Thanks!

[@ภพภูมิ-ผ8ฌ]

This is the best way to teach " How to think" by mathematic imagine. Very good !

[@dpi168]

Thank you, You make math so much fun!

[@niburX]

x⁴ + 64 = 0 x⁴ = -64 x² = ±sqrt(-64) x² = ±8i x² = 8i, -8i x = ±sqrt(8i), ±sqrt(-8i) x = ±(2+2i), ±(2-2i) x = 2+2i, -2-2i, 2-2i, -2+2i

[@Masteroogway-e4i]

I was thinking the same

[@АлексейБигвава]

Все верно. Но в его случае очень красивое и понятное геометрическое обоснование. Оно позволяет по иному взглянуть на алгебраическую запись.

[@Masteroogway-e4i]

@@АлексейБигвава English please

[@АлексейБигвава]

@Masteroogway-e4i You are right. But his geometry explanation of solution is so clear. Very understandable.

[@Masteroogway-e4i]

But- aren't we supposed to do it the easier way ?

[@rbarnes4076]

@bprp math basics Love your channel! Super-informative, even if you've been doing math for decades!

[@bendunselman]

How about quaternion or octonion solutions?

[@marco5w]

Use complex conjugate numbers! a2 + B2 = (a+bi) * (a-bi) a=x 2 b=8 and solve two simple quadratic eq.

[@novemtrigintillionaire7684]

If you think about it, plus-minus is just a specific case of the roots of unity where you raise it to fhe second. We could just introduce terminology thatg adds a aspecofic number to denote that set, but its somewhat complicated

[@Chriib]

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.

[@dogbreaththe3rd851]

64 = -64i^2 substitute giving x^4 - 64i^2 The factorization then becomes a difference of squares (x^2 -8i) * (x^2 + 8i)

[@noxfelis5333]

There is something funny you can do with this special case. where x^4 = - y Since: (a+ai)^4 = (a^2 + 2a^2i - a^2)^2 = - 4a^4 , (a-ai)^4 = (a^2 -2a^2i - a^2)^2 = - 4a^4 and for negatives you can just take the -1^4 out and turn it into 1 Put in y = 4a^4 you can then just do square(square(64/4)) and then put in the answers in place of a and get all 4 answers with just changing the signs before a. square(square(64/4)) = square(square(16)) = 2 making x = 2 + 2i , x = 2 - 2i , x = -2 + 2i , x = -2 - 2i

[@Galileosays]

Another way is by taking x=a+bi with a,b are real numbers. From a^4+4a^3bi-6a^2b^2-4ab^3i+b^4+64=0 follows: +4a^3bi-4ab^3i=0 => a=+/-b. and from substitution of this in a^4+-6a^2b^2+b^4+64=0 4a^4=64 gives a=+/-2. Hence x=+/-2+/-2i.

[@janda1258]

Just multiply by 1, e^(n*2pi). The n takes care of all roots when taking any root of a number

[@candi_ositos]

The first thing i thought was to subtract 64 on both sides, then square root both sides without forgetting the plus or minus, then square root both sides without forgetting the plus or minus and the answer is ±2*sqrt(±2i)

[@rob876]

x^4 = 64e^(iπ(1+2n)) x = 4e^(iπ(1+2n)/4) x = 4e^(iπ/4), 4e^(3iπ/4), 4e^(5iπ/4), 4e^(7iπ/4) x = (2 + 2 i)√2, x = (-2 + 2 i)√2, x = (-2 - 2 i)√2, x = (2 - 2 i)√2

[@noxfelis5333]

4^4 is 256 not 64, which is 4 times as much, remove √2 from the final answer and place in √8e^(iπ(1+2n)/4) instead of 4e^(iπ(1+2n)/4)

[@maciejkubera1536]

Actually if You have a square of x equal to some number a, the x can be + or - square root of a, or, to put it the other way, the solutions are the the numbers obtained by multiplying square root of a by square roots of unity. Almost the same is for the x to the fourth power equal to a, but you take all fourth roots of unity and multiply it by the fourth root of a. That’s why writing plus or minus is not the whole answer. Am I wrong?

[@dlevi67]

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.

[@Taric25]

Ha! Good luck getting an algebra student to take the square root of 8i

[@dlevi67]

@@Taric25 Why? What is the problem? Teach them to convert it to polar form (all stuff that "I" did at 16) and it's very easy. If you don't want to do that, you can still use the "complete the square" method for the two degree 2 factors.

[@Taric25]

@@dlevi67 Polar form isn't taught until after pre-calculus. I didn't learn it until Calculus II. Completing the square is absolutely what you should teach an algebra student.

[@dlevi67]

@@Taric25 Your "absolutely" is far, far less absolute than you believe. You keep assuming that the whole world learns maths following the same syllabus that you did. That is simply not true. As I said, we were taught complex numbers in polar and rectangular form at 16.

[@Taric25]

@@dlevi67 I have never heard of complex numbers in polar form taught in high school but good for you.

[@lornacy]

Does the Po-Shen Loh method work with a quartic like this?

[@donmoore7785]

Someone above did that in the comments.

[@hydo-7653]

Amazing video! this is love from the 8th graders haha

[@martincohen8991]

More generally, x^4+4n^4=(x^2+2n^2)^2-(2xn)^2=(x^2+2n^2-2xn)(x^2+2n^2+2xn).

[@felixparley]

Just use (64*e^(i*180))^(1/4) = 64^(1/4)*e^(i*(180/4 + j*(360/4))) for j = 0..3 = 2*sqrt(2)*e^(i*(45+90j)) for j=0..3

[@RyanLewis-Johnson-wq6xs]

Input (-2 - 2 i)^4 + 64 = 0 Result True

[@bobbun9630]

Completing the square seems unnecessarily complicated for this one. Since you know you're looking for complex solutions, just use the fact that a sum of squares is always factorable over the complex numbers from the outset. More explicitly, just pull an i^2=-1 out of the 64 to convert it into a difference of squares, and proceed from there.

[@leikang252]

Mr. Teacher, I am wondering where you did learn specialized English in maths? Amazing.

[@ВалерийХарченко-ш5д]

4 roots: x=(+-sqr(+-sqr(-64))

[@leonardobarrera2816]

This video must to be called, How to demostrate the sqrt(i) to non caclulus studients!!!

[@alexandermorozov2248]

The root of i can be easily represented graphically on the complex plane 😃

[@leonardobarrera2816]

@ yeah, but this was done by algebra That is awesome!!!

[@leonardobarrera2816]

@@alexandermorozov2248 but is not easy for algebra students

[@chinamensuki8170]

a^4+4b^4=(a^2＋2ab＋2b^2)(a²－2ab＋2b^2)

[@darylcheshire1618]

I plotted it and got a flat-ish parabola intersecting the y axis at 64.

[@EduardoSousaSaraiva-p6i]

2*sqrt(2)*exp(i(pi/4 +2*pi*k)) k element of Z

[@stpat7614]

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)

[@sina4948]

Why ±√-64 is not amongst the answers?

[@NLGeebee]

Because 1) it is a 4th power 2) calculating with ⁴√-64 to directly ±2 ±2i is sloppy maths ;)

[@donmoore7785]

Because that is not in simplest form. You need to find out what √-64 is.

[@ugurcansayan]

cis to the rescue x = |2√2| · cis (45° + k·90°) (i rotates numbers 90°, i² rotates numbers 180°, i³ rotates numbers 270°, therefore i^½ MUST rotate numbers 45°)

[@RyanLewis-Johnson-wq6xs]

(-2-2i)^4+64=0 x=±2±2i It’s in my head.

[@excentrisitet7922]

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)

[@Taric25]

Because there is no way an algebra student would know complex analysis, obviously

[@wristdisabledwriter2893]

Wasn’t this method figured out by my favorite mathematician Sophie Germain?

[@APerson-ws4cw]

I'm usually a math wizz, this was wild to witness lol

[@ivangospodarski80]

Let x²=y, y=±8i, then we solve for x²=8i, x=±2√2i, and then for x²=-8i, x=±2i√2i

[@tomtke7351]

x^4 => REQUIRES 4 solutions

[@Nazimİsmayılov-e9u]

Bu məsələ Sofi Jermen məsələsinin dəyişdirilmiş şəklidir.Sofi Jermen məsələsində X-in 1-dən fərqli natural qiymətlərində X^4+4 ədədinin mürəkkəb ədəd olduğunu isbat etmək tələb olunur. Bu məsələdə 4 əvəzinə 4 ilə istənilən həqiqi ədədin 4-cü dərəcədən qüvvətinin hasilini yazıb tənlik kimi həll etdikdə tənliyin kökləri həmişə kompleks ədəd olur.Bu məsələdə 4 əvəzinə 4*2^4=4*16=64 yazılıb. Təşəkkürlər.

[@Skyler827]

I would have plugged in x = Re^(i×theta)

[@F1r1at]

x^4 = -64 x^2 = +-8i x1 = sqrt(8i) = 2*sqrt(2)*sqrt(i) = 2*srqrt(2)*((1+i)/sqrt(2)) = 2 + 2i x2 = -sqrt(8i) = x1*(-1) = -2 - 2i x3 = sqrt(-8i) = (2+2i)*i = 2i+2i^2 = 2i - 2 = -2 + 2i x4 = - sqrt(-8i) = x3*(-1) = 2 - 2i

[@CasiMediocre]

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"

[@trueriver1950]

I think that's a valid destination of the same answer

[@fastneuro9829]

Ferrari formula

[@MichaelRothwell1]

Overkill

[@cubetoast976]

So, basically using the Sophie Germain identity

[@alucardthespy5539]

x^4 + 64 = 0 x^4 = -64 sqrt(x^4) = [+/-]sqrt(-64) x^2 = [+/-]8i x^2 = 8i ; -8i sqrt(x^2) = [+/-]sqrt(8i) ; [+/-]sqrt(-8i) x = [+/-](2 + 2i) ; [+/-](2i - 2) x = 2 + 2i ; -2 - 2i ; 2i - 2 ; 2 - 2i OR... x = [+/-]2 + [+/-]2i

[@henkhu100]

Again he makes an error he often makes: At 10:20 he replaces √(x+2)^2. by x+2. (by erasing √ and the exponent 2) But that means that for instance √(-3)^2 is equal to -3 and that's wrong because √ stands for the principal square root and √(-3)^2 = √9 = 3 Skipping the √ symbol against the exponent 2 is an often made error.

[@CR7inUefaCristianoLeague]

this is taught for us in grade 8 in india this is kind of simple only different thing is quadratic formula but its kind of simple maybe its just me solving too many questions but this is common logic with geometry as well thanks for the recap anyways I appreciate the content

[@HarshvardhanMishra-ql2mj]

Good one❤

[@marceliusmartirosianas6104]

(x^4+ 60^4)^o=O]=[ x^4+60^4]^1=1 ]=[ x^4= 1- 60^4 ]=[ 1/60^4]=[ 60^4=1=60^4-1= 60^4/1=6o^4 ]=[X^4=60^4]=[ X=60 }]=== ACADEMIC MARCELIUS MARTIROSIANAS

[@dogbreaththe3rd851]

64 = -64i^2 substitute giving x^4 - 64i^2 The factorization then becomes a difference of squares (x^2 -8i) * (x^2 + 8i)

[@ricardoguzman5014]

X⁴+64=0 X⁴-(-64)=0 X⁴-(8i)²=0 (X²-8i)(X²+8i)=0 1. X²-8i=0 X²-4(2i)=0 X²-2²(1+i)²=0 (X+2(1+i))(X-2(1+i))=0 X+2(1+i)=0 or X-2(1+i)=0 X= -2(1+i) or X =2(1+i) X= -2-2i or X =2+2i 2. X²+8i =0 X²-(-8i)=0 X²-(-4(2i))=0 X²-(2i)²(1+i)²=0 (X+(2i(1+i))(X-(2i(1+i))=0 X+2i(1+i)=0 or X -2i(1+i)=0 X+2i +2i²=0 or X-2i-2i²=0 X+2i-2 =0 or X-2i+2=0 X=-2i+2 or X=2i-2 The four solutions are: X= -2-2i or X= 2+2i or X= -2i+2 or X=2i-2

[@Hunni125]

makes sense

[@Xlrupt]

Looks good

[@binglefish_6742]

So much easier to do x^4 = exp(pi.i+2n.pi.i) and 4th root that

[@fifabuilder23]

2.83 is also right

[@ChavoMysterio]

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 ❤❤

[@dogbreaththe3rd851]

x^4 - 64i^2 = 0 is easier - difference of squares requires no geometry

[@Khusbuhasrat]

Wowwwwwwww i luv geometry solution.. Really u r genious

[@donmoore7785]

You missed the opportunity to point out expicitly that a^2 + b^2 = (a + b)(a + b) - 2ab, when you showed it graphically.

[@cursosGT]

It's very Easy by Ruffini method..

[@Videogamewrestling22]

Thank you daddy

[@BN-hy1nd]

Brilliant😅

[@tedr.5978]

11 minute video to get the same 4 complex number solutions you get in 11 seconds by taking the cubed root of both side of the original equation.

[@kay5718]

This is for students learning algebra, if you already know how to do this the "better" way, it's not for you

[@tedr.5978]

​@@kay5718 You say this is for students learning algebra. But that is not what the video says. Starting at 0:30, referring to the easy, quick and straight forward way to solve the equation with algebra, the video states "Please don't do like this. Because in fact for the equation you are not going to get any real roots." This fausly implying the long solution will find real roots the short solution will not. The point of teaching algebra to students is so they can use it to solve real problems in science, engineering, economics, etc. Math for math's sake may give math teachers an hard on, but it does not help anyone else. Students should do the work to understand derivations, but once they know them, don't point to the better method that gets the same correct answer (like this video does at 0:24) and say "please don't do this."

[@Danker1248]

I was ablw to solve a different way x^4 + 64 = 0 x^4 = -64 x^4 = (-4)^3 x^(4/3) = -4 (4/3)lnx = ln(-4) (4/3)lnx = ln(4) + ln(-1) (4/3)lnx = ln(4) + ln(e^(iπ)) (4/3)lnx = ln(4) + iπ lnx = (3/4)(ln(4) + iπ) x = e^(((3/4)ln(4)) + (3/4)iπ) x = e^((3/4)ln4)) * e^((3/4)iπ) x = e^((3/4)ln4)) * (i * sqrt(2)/2 - sqrt(2)/2) Not a very nice looking answer but plugging it into desmos will show it checks out, lol

[@BogdanTestsSoftware]

I came here for the E joke on the Tshirt, :( came out disappointed. Here's a joke to make up for it, until we find a better one: perhaps he should be used as a source of logs, b/c no matter how much you log him, he's still the same. I could not find an epsilon joke that would matter.

[@alexandermorozov2248]

If you solve the equation in the traditional way, the answer will be the same: sqrt(sqrt(-64))=±2±2 i 😅

[@carultch]

With the help of DeMoivre Whose theorem we love ya There's fourth roots all over the plane Yes they're complex But do not perplex A new kind of numbers we gain

[@raymondarata6549]

DeMoivre is de man. That's how I did the problem.

[@carultch]

@@raymondarata6549 That's the final stanza to my poem about the cubic formula.

[@alwayschill4522]

x^4=-64 x^2=+- 8i x= +- sqrt (+- 8i) wait a second... what is the square root of plus minus 8i 😭

[@alawiwatermelon4380]

x=±2±2i

[@pavulugjimene8869]

(-2) =X X^4 = X*X*X*X 2^4 =2*2*2*2 2*2=2+2 2*2*2*2= (2+2)+(2+2)+(2+2)+(2+2)+(2+2)+(2+2)+(2+2)+(2+2)+ (2+2)+(2+2)+(2+2)+(2+2)+(2+2)+(2+2)+(2+2)+(2+2)=32*(2) X*X*X*X= (X+X)+(X+X)+(X+X)+(X+X)+(X+X)+(X+X)+(X+X)+(X+X)+(X+X)+(X+X)+(X+X)+(X+X)+(X+X)+(X+X)+(X+X)+(X+X)=32*(X) X^4= 32*(-2)=-64 🤣

[@slava6105]

But Mouivre's formula...

[@SarojtapashLalita]

Explain bro 🎉

[@slava6105]

@SarojtapashLalita There are already noticeable comments with solutions based ln Moivre's formula. My comment is a geometric interpretation of what I've learned from calc 1 in university. The idea of it is that all solutions of z^n = r^n (z and r are compex, n is integer) are evenly spaced on circle centered at (0;0) with radius |r|. Knowing that there are a total of n solutions to n-degree complex polynomial, and that they all have the same radius and are evenly spaced, we can find: 1) one starting solution, 2) angle at which every next solution will be placed. Moivre's formula from wikipedia: (cosx + isinx)^n = cos(nx) + isin(nx) (Euler's formula) exp(ix)^n = exp(inx) Our university's book formula derived from them: z^n = a; φ = Arg(a); r = |a|^(1/n) Arg is angle of number on complex plane z_k = r*exp(i*(φ/n + 2πk/n)), k=0..n-1 φ/n is a starting angle and 2π/n is step size. Given x⁴ + 64 = 0 => x⁴ = -64 First, find radius: r = |-64|^(1/4) = 64^(1/4) = 2^(6/4) = 2√2 Then, find starting angle: Since -64 is on negative real side, φ = π. With other values I would have to go through some trigonometry with real/imaginary parts. Therefore φ/n = π/4. And step angle is 2π/n = 2π/4 = π/4. So, we start at (2√2)*exp(i*π/4) and rotate +π/4 (ccw) each time. z0 = 2√2*exp(i*π/4) z1 = 2√2*exp(i*(π/4+1*π/2)) = 2√2*exp(i*3π/4) z2 = 2√2*exp(i*(π/4+2*π/2)) = 2√2*exp(i*5π/4) z3 = 2√2*exp(i*(π/4+3*π/2)) = 2√2*exp(i*(7π/4)) Leaving it in polar form since converting them back is trgonimetry that is not mine. Desmos says they are: 2+2i, -2+2i, -2-2i, -2-2i respectively.

[@maxime9636]

🙏🏻🙏🏻🙏🏻❤️❤️❤️👍🏻👍🏻👍🏻all in geo

[@praporserg]

Для чего вообще нужны такие уравнения? Никакого от них практического смысла. Очевидно, что такое уравнение не будет равно нулю при любых х. Зачем вводить число i, не имеющие никакого физического смысла, чтобы потом решать бесполезные уравнения. 4 + х^2 =0 Допустим,что на самом деле х^2 это - х^2, тогда х = ±2. В чем смысл такого решения, никто же не просил посчитать уравнение 4 - х^2 = 0.

[@xavariusquest4603]

No.

[@-Yousof-]

noice! 2nd comment btw

[@GillesF31]

Your reasoning reminds me of a remarkable identity I used (see below) to solve the equation in a path that appears to be identical to yours... x⁴ + 64 = 0 (x²)² + 8² = 0 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ | recall: a² + b² (with 2ab = c²) = (a + b - c)·(a + b + c) | ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ • a = x² • b = 8 => 2ab = 16x² = c² => c = 4x => (x²)² + 8² = 0 becomes: (x² + 8 - 4x)·(x² + 8 + 4x) = 0 --- /// case: (x² + 8 - 4x) = 0: (x² + 8 - 4x) = 0 x² - 4x + 8 = 0 Δ = (-4)² - 4·1·8 = 16 - 32 = -16 √Δ = ±i√16 = ±4i • root #1: x = (-(-4) + 4i)/(2·1) = 2 + 2i • root #2: x = (-(-4) - 4i)/(2·1) = 2 - 2i --- /// case: (x² + 8 + 4x) = 0: (x² + 8 + 4x) = 0 x² + 4x + 8 = 0 Δ = 4² - 4·1·8 = 16 - 32 = -16 √Δ = ±i√16 = ±4i • root #3: x = (-4 + 4i)/(2·1) = -2 + 2i • root #4: x = (-4 - 4i)/(2·1) = -2 - 2i --- /// final results: ■ root #1: x = 2 + 2i ■ root #2: x = 2 - 2i ■ root #3: x = -2 + 2i ■ root #4: x = -2 - 2i 🙂

[@fernandocardenaspiepereit4097]

It has no solutions, Imaginary numbers don't exist.

[@guidomista8448]

They kinda do exist.

[@donmoore7785]

Oh dear...

[@perlman7376]

@@guidomista8448 Give me a value for X (only 1 value) that fits the equation X^4 + 64 = 0. Math slight of hand will not do it.

[@guidomista8448]

@@perlman7376 there are no real solutions, if that's what you're asking. But if we go into the set of the complex numbers then there is a solution.

[@elmanmusaiev7003]

There is no solution for this on rational side.

[@doanhnguyenquoc4707]

Bài toán vô nghĩa!!!