How to solve (x+2)^4=x^4+2^4 besides x=0
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- เผยแพร่เมื่อ 11 ก.ย. 2024
- The equation (x+2)^4=x^4+2^4 is a seemingly false identity but it's actually a nice algebraic equation. It has not just x=0 as a solution but two more complex solutions!
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More practice with expanding binomials: th-cam.com/video/WzF_UWQ-bEE/w-d-xo.htmlsi=PaIZ4vaYtEtFS0ph
binomial expansion theorem i thought, but that is a small distinction.
Wow! I love the use of Pascal’s triangle there. It made everything else so much easier.
you can use Combination too, nCr
It's combinatorics
I was with you until everything you said after x=0
He just set the other factor equal to 0 which is a quadratic that can't be factored so he used the quadratic formula.
So you understood the first 28 seconds. You weren't with him for very long.
@@chitlitlah no, admittedly not.
@@A_Stereotypical_Heretic No sweat. Watch enough of these videos and you'll become a math guru.
(commented before watching)
When you see that x=0 is a solution, there will remain a quadratic equasion you can solve.
2:15 I always put the 0th powers down, it cements the pattern. I mean, you don't need to put the exponents of 1 down either but you did, the same logic applies.
The Binomial Theorem states:
\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]
I got the same thing, just multiplied by 2 because I didn’t divide 2 on both sides.
I was expecting 4 solutions because of the 4th power in the initial equation….🤔
That's a good intuition. This time the fourth power is cancelled away. Like, we know x²=x+1 has two solutions, if I add x²² to both sides, I'm not making twenty solutions to magically appear. :)
I guess zero is a repeated root (solution) so all of the solutions are 0,0 and the two non real solutions
It's actually just a cubic equation because there is x^4 exactly once on both sides
@@sreenidhiyamana231That would be the case if the polynomial is divisible by x^2, which is not the case
@@antoniusnies-komponistpian2172 that's quite a good answer
I have a question if (a+b)^2 can be expanded so lets say a+b = x so what will be x^2( i have a formula for expanding x^2 note: its not x*x) if u can respond i ll be happy
It's true in a field of characteristic 2
Average numberphile viewer be like:
Am I missing something? (X + Y)^2 = X^2 + Y^2 + 2XY
It can never be true?
Sorry in advance if I asked a dumb question
That is always true. Which is why it's called an "identity" instead of just an "equation". An identity is something akin to a law, in that it is universally true for all values of the variable(s).
Here, (x+2)⁴ = x⁴+2⁴ is an equation. It is NOT true for all values of x. We have to solve the equation to get the values of x for which this expression is true. One of such values, as explained at the start of the video, is zero.
Hope that helped :)
complex solutions do matter
I love this guy
*should* separate the rational to 2 terms to get into the a+bi form and have I after the radical of course to match the Complex number form.
Wanna share this technic I love for the last equation ;
x²+3x+4 = 0 = (x+1.5)²+1.75 (1.75=7/4) (x+1.5)²-(√(7/4))² = 0 so the solutions are 1.5±i√7/2
Just for generalization, x²+bx+c = 0 => (x+b/2)²+(c-(b/2)²) = 0 => (x+b/2)²=((b/2)²-c) = > x+b/2=±sqrt((b/2)²-c) => x = -b/2 ± sqrt((b/2)²-c) => x = -b/2 ± sqrt(b²/4 - c) = -b/2 ± sqrt((b²-4c)/4) = (-b±sqrt(b²-4c))/2. You just proved quadratic formula :D
Obs: in this case i assume a equation linearly simplified to 1x².
@@Andre-he5ly you're right
I think that on the ring R=Z/4Z the equality holds for every x in R...
This was actually EXACTLY the method I used to solve this.
Lo hubieras resuelto más fácil con diferencia de cuadrados
Bprp, there's another possible solution which I guess is not in our level, as degree = 4
Binomial theorem makes things nice
It seems that the answer to any "does this have any solutions?" is always "yes, complex solutions".
That's not always the case.
x+1 = x does not have any solution, even in the complex world.
Nice. Cool
i ❤ Mathematics
The X values are: 0, -1.5+((sqrt(7)i)/2), and -1.5-(sqrt(7)i/2)
Isn't /2/2=/4
@@massivecowbreakout7555my bad
not exactly correct.See video.
@@whoff59 how?
@@yaseenelhosseiny you forgot the imaginary unit
its i*sqrt(7) not sqrt(7), so -1,5 + i * sqrt(7)/2 and -1,5 - i * sqrt(7)/2
the extra pair of parentheses around the sqrt is unnecessary
😺🫵
O is only real soln.... Why to make such drama
Cause math fun
I've been trying to decide how to respond to this, and I can think of several possibilities:
1. You have a very different idea of "drama" than I do if you think that calmly working through the solution to a math problem is "drama."
2. If you didn't care, why did you choose to watch the video?
3. How do you know that 0 is the only real solution without working through the problem?
4. "Real" (and "imaginary") are technical terms in math. Don't assume that non-real numbers don't matter or aren't important. (Whether they matter or not in a particular prpblem can depend on the context.)
5. All the math in this video is pretty standard stuff that any algebra student at a sufficiently high level should be able to handle.
@@Steve_Stowersyou likely just replied to a tongue-in-cheek comment with an itemized list. Lmao. I'm pretty sure they meant drama as in all the extra work of finding complex solutions, not emotional conflict. Per your fifth point, unless the unit is about complex numbers specifically, most of the time teachers will only care that students find the real solutions.
@@albericponcedeleon2696 No they won't. I don't know where you are from, but teachers always ask for complex solutions after they have taught you the complex numbers chapter. And Steve_Stowers every point is correct and reasonable unlike you.
@@albericponcedeleon2696 even if that's the case, you still need to do the rest to make sure you're not missing any real solutions