Solving an Undecic Polynomial Equation
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- Опубліковано 2 жов 2024
- This video is about a rational equation that can be solved with an algebraic trick.
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The equation looks scary first but then it slowly unfolds. Don't be discouraged just because something looks intimidating.
Wow That Was Slick!!!
Ωραια σκεψη!!!
I did what you said would not work - dividing numerator and denominator by x - and it worked quite well!
We get (x¹⁰+1)/(x⁶+x⁴) = 206/16
We have only even powers of x, so substitute y=x²
(y⁵+1)/(y³+y²) = (y⁵+1)/(y²(y+1)) = 205/16
Notice that y⁵+1 = (y+1)(y⁴-y³+y²-y+1) (This works for any odd power plus one)
( y+1)(y⁴-y³+y²-y+1)/(y²(y+1)) = 205/16
Cancel out (y+1)
(y⁴-y³+y²-y+1)/y² = 205/16
Divide through by y²
y²-y+1-y⁻¹+y⁻² = 205/16
Now substitute u = y+y⁻¹
Then we get u² = y²+2+y⁻² => y²+y⁻² = u²-2
So we get
u²-2+u+1 = 205/16
Collect all terms to one side
u²+u-221/16 = 0
Solve for u
u = 17/4 or u = -13/4
Solve for y
y+y⁻¹ = 17/4 = 4¼ => We see directly that the solutions are y = 4 and y = ¼
y+y⁻¹ = -13/4 => y = (-13±sqrt(105)) / 8 These two roots are negative
Solve for x, where y = x²
y = 4 => x = ±2
y = ¼ => x = ±½
The negative y values yield no real solutions for x
And that's it!
Nice!
Awesome formatting!
I had a similar idea that led to the same end:
(x^11+x)/(x^7+x^5)=(x^5+1/x^5)/(x+1/x) (we divide everything by x^6).
After that we notice that if x=-1/x, x^5=-1/x^5. Therefore we know that we can factorise by (x+1/x).
x^5+1/x^5=(x+1/x)(x^4-x²+1-1/x²+1/x^4)
Our equation becomes: x^4-x²+1-1/x²+1/x^4=205/16.
If y=x² we have: y²-y+1-1/y+1/y²=205/16. (1)
(y+1/y)²=y²+1/y²+2
Then (1) (y+1/y)²-(y+1/y)-1=205/16
If z=y+1/y then (1) z²-z-221/16=0.
Delta=1+221/4=225/4=(15/2)²
The solutions are 17/4 and -13/4.
After that we have to solve:
y+1/y=17/4 (2) and y+1/y=-13/4 (3)
(2) y²-17/4.y+1=0
Delta=289/16-64/16=(15/4)²
The solutions of (2) are (17/4-15/4)/2=1/4 and (17/4+15/4)/2=4.
(3) y²-13/4.y+1=0
Delta=169/16-64/16=105/16=(sqrt(105)/4)²
The solutions of (3) are (-13+sqrt(105))/8 and (-13-sqrt(105))/8
Finally, for each solution of (2) and (3) we have to solve x²=y.
For the solutions of (3) we obtain 1/2, -1/2, 2 and -2.
For the solutions of (4) we have negative values so we have complex solutions: i.sqrt(2)/4.sqrt(sqrt(105)-13)), -i.sqrt(2)/4.sqrt(sqrt(105)-13)), i.sqrt(2)/4.sqrt(sqrt(105)+13)) an -i.sqrt(2)/4.sqrt(sqrt(105)+13)).
molt be
Solid......bro.👍
My take on the quartic equation at 12:10:
16u^4 - 80² - 125 = 0
=> (4u²)² - 20(4u²) - 125 = 0
If we let t = 4u²:
=> t² - 20t - 125 = 0
Using the method used by 3B1B in his first episode of Lockdown Math (ua-cam.com/video/MHXO86wKeDY/v-deo.html), we can easily solve this quadratic equation:
m = -20/2 = 10 and d = sqrt(10² - (-125)) = sqrt(225) = 15,
and thus:
t = m +- d = 10 +- 15 => t = 25 or t = -5
As t = 4u² which is positive, the last solution can be discarded.
Finally:
t = 4u² = 25 => u² = 25/4 => u = +- 5/2
What do you think? I think the procedure you came up with was rather lengthy and not something I'd be able to see myself.
Nice! I like it!
If ax^2 + bx + c = 0 and b is even, there is special formula: x = (-b/2 ± SQRT(b/2-ac))/a. So y = (40 ± SQRT(40^2 +16*125))/16
Yes! Thank you!
Actually with this method you can find all of ten solution (as this equation is tenth-degree, because we exclude x=0), because from x+1/x=0 you have i and -i as solution, and from u^2=-40/32 you have another four complex solution)
Nice!
But if we put i or -i back in the original equation, you don’t get 205/16. So that Dan not necessarily be a solution. Also he put real solutions in the thumbnail anyway so. Correct me if I’m wrong
-40/32=
-5/4
Heart filled with joy . Interesting method of solving .God bless you!
So nice of you! 💖
And if you graph (16(x¹¹+x)/(x⁷+x⁵))-205=y you get a somewhat eccentric curve nicely crossing the y=0 line at x={-2;-0.5;0.5;2}
before following the video for the solutions, I tried some values for x. for x=2 , then: x**7 +x **5 = 160 . and x**11 + x**1 = 1050; so 1050/160 = 105/16 therefore x=2 is a solution. Likewise, remembering another similar problem from another site, I tried x= 1/2, and after some basic operations , there came thee same result. so I got 2 solutions: x=2 and x=1/2. anyway it could cost time to find out whether these are the only two solutions.
No tricks other than factorization and the rational root theorem are required to solve that equation. Begin by canceling out an X from numerator and denominator to get
(X^10 + 1) / (X^4 * (X^2 + 1))
and then notice that X^10 + 1^10 factors as (X^2 + 1^2)*(X^8 - X^6 + X^4 - X^2 + 1) which is immediately obvious if you remember that A^5 + B^5 factors as (A + B)*(A^4 - A^3*B + A^2*B^2 - A*B^3 + B^4) after you rewrite X^10 + 1 = (X^2)^5 + 1^5
Now X^2 + 1 cancels in the numerator and denominator, leaving
(X^8 - X^6 + X^4 - X^2 + 1) / X^4 = 205/16
But we can substitute Y = X^2 and rearrange to standard form to get
16*Y^4 - 16*Y^3 - 189*Y^2 - 16*Y + 16 = 0
which could be further reduced with the right trick but I prefer to avoid tricks unless necessary. Not necessary here since the rational root theorem gives us plus/minus 1/16, 1/8, 1/4, 1/2, 1, 2, 4, 8, 16 to try. It turns out that Y = 4 and Y = 1/4 are roots, so that factors to:
(Y - 4)*(4*Y - 1)*(4*Y^2 + 13*Y +4) = 0
The quadratic does not have any positive roots, and since X = sqrt(Y), the solutions we get are sqrt(4) and sqrt(1/4) or 2 and 1/2.
What you do looks like magic. I would appreciate much more if you motivated what lead you down this path. It was not an obvious approach at all from where I am sitting.
After rewrite above equation as
(x^10+1)/(x^6+x^4)=(205k/16k)
From this We get 2equations
First x^10+1=205k
x^6+x^4=16k
x^4(x^2+1)=16k
From second equation
We have x is even and k is odd.
(x^11+x)/(x^7+x^5)=(205/16)
16(x^11+x)=205(x^7+x^5)
16x^11+16x=205x^7+205x^5
16x^11-205x^7-205x^5+16x=0
x(16x^10-205x^6-205x^4+16)=0
x=0 is an extraneous solution since the denominator contains the variable x. That's as far as I can go.
At 1:18 dividing by x is ok. (X^10+1 )/(x^6+x^4} = 205/16=A
. (X^10+1 ) =A (x^6+x^4} divide by x^5
x^5+1/x^5 = A ( x + 1/x)= Au. and as before, x^5+1/x^5 =u^5 - 5u^3+5u=Au
y^2 -5y =A-5= 25/16
At 12:15 y=4u gives y^2 -20y-125=0 which makes it easier to spot factors of 5 and 25
Giving (y-25)(y+5)=0 So, 4u^2 = 25,-5. By the way it would be good if you could freeze the top right corner of the frame so you could keep the question and substitutions and lemmas.
16y^2 - 80y - 125 = 0
The equation would need to have two numbers that are 20 apart which are 25 and 5 and the larger would have to be negative, provided that 16 has a perfect square root. This yields:
(4y - 25 ) ( 4y + 5 ) = 0
Good thinking!
@@SyberMath i am glad that you approve.
You can also do the following: Multiply 16 by -125 to get -2000. Find two numbers whose product is -2000 and whose sum is -80. Those numbers are -100 and 20. Then write 16y^2 - 80y - 125 = 0 as 16y^2 - 100y + 20y - 125 = 4y(4y - 25) + 5(4y - 25) = (4y - 25)(4y + 5) = 0
@@SyberMath very nice but it is difficult to fond factors of large numbers like that
@@SyberMath also i found formula based on a, b, and c values of quadratic equation that will give the two numbers to multiply to one number and add to another.
if ax^2 + bx + c = 0 then the two numbers are :
[ b + sqrt ( b^2 - 4ac ) ] / 2 and
[ b - sqrt ( b^2 - 4ac ) ] / 2
It is logical to check equality if the denominators are the same, x^5=16x. Obviously, 2 and -2 are roots. Since x and 1 / x have equal rights in the formula, 1 / x = ± 2 are also roots. However, other roots need to be explored and this is not so fast. I used factorization. Knowing the roots makes the task easier.
LS=(x^10+1)/(x^4*(1+x^2))=((x^2)^5+1)/((1+x^2)*x^4).There Is substitution y=x^2. Then( (y^5+1):(y+1))=(y^4-y^3+y^2-y+1=x^8-x^6+x^4-x^2+1. We have then
LS=x^4-x^2+1-1/x^2+1/x^4, there is geometr series So (1/x^6+1):(1/x^2+1)=205/16. After substitution 1/x^2=z we have (z^3+1):(z+1)=205/16, So z^2-z+1=205/16
Aslında x^2=a dönğşümünden sonra
hiçbir şey kalmıyor.
Here is a better trick:
Define Z(n):= X^n + X^(-n); (Z(o) = 2)
Z(n) satisfies recursion formula:
Z(n+1) = Z(n)*Z(1) - Z(n-1); ( Also (Z(2n) = Z(n)^2 - Z(o))...(i)
Problem
125/16 = (x^11 + x)/(x^7 +x) = Z(5)/Z(1) (i.e just divide by x^6)
from recursion formula (i)
Z(5)/Z(1) = Z(1)^4 - 5*Z(1)^2 +5 = 125/16 ...(ii)
Solving (ii) gives Z(1)=5/2, Z(5)= 1025/32 (=(125/16)*Z(1))
For info [Z(1)..Z(5)] = [5/2, 17/4, 65/8, 257/16 1025/32]
A bit of work gives
(x^5 + 1/x^5) /(x+ 1/x) = 205/16
Again (x^2 + 1/x^2) = (x+1/x) ^2 -2
& (x^3 + 1/x^3) = (x+1/x)^3 -3(x+1/x)
Multiplying these two identities:
x^5 + 1/x^5 + x + 1/x equals to
(x+1/x)((x+1/x) ^2 -3)((x+1/x) ^2 -2)
Therefore
(x^5 + 1/x^5) /(x+ 1/x)
=((x+1/x)^2 -3)((x+1/x)^2 -2) -1
= u^2 - 5u +5
Here in u represents (x+1/x)^2
Hereby given equation becomes
u^2 - 5u = -5 + 205/16
or 16 u^2 - 80 u -125 = 0
or (4u - 25)(4u + 5) = 0
case 1 : (x+1/x) ^2 = u = 25/4
or (x+1/x -5/2) (x+1/x +-5/2) = 0
or (2x^2 -5x +2) (2x^2 -5x +2) = 0
or (2x -1) (x-2) ( 2x+1) (x+2) = 0
or x = -2, -1/2, 1/2, 2
Case 2 : (x+1/x) ^2 = u = -5/4
or x^2 + 1/x^2 = - 13/4
or 4 x^4 + 13* x^2 + 4 = 0
No solution in real domain
Eyeballed the original rational equation, the top looked suspiciously like 2050 = 2048 + 2 = 2^11 + 2, and indeed that is a solution. Then x = -2 is immediately also a solution either by
1) the fact the rational expression has all odd powers in both numerator / denominator, means it will have a -1 in both numerator / denominator.
2) cross-multiplying and dividing by x (x = 0 is inadmissible solution), then using the fact the resultant polynomial 16x^10 − 205x^6 − 205x^4 + 16 has only even powers.
x = ±1/2 as a guess from the Rational Root Theorem is easy to verify as a root, and dividing by (x^2 − 4)(4x^2 − 1) leads to yet another symmetric polynomial 4x^6 +17x^4 + 17x^2 + 4 which can be reduced to 4z^3 +17z^2 +17z + 4 (which has z = −1 as a root) by substituting z = x^2.
From the above the complex roots are ±i, ± i /(2 sqrt(2)) * sqrt(13 ± sqrt(105)), the latter just being a shorthand to write 4 roots as it leads to (+, +) (+, −) (−, −) (−, +), where the first part of the tuple is the sign of i, and the second is the sign of sqrt(105)
Definitely one of the neatest 10th or 11th-degree polynomials out there!
Wow, impressive 😉
So were you able to see in advance that this method would actually work, or was it just hard work a la " just try it and see what happens?" - or just sheer luck? 🤔😂
pull x out . then multiple every terms will get 16x^10-205x^6-205x^4+16 =0 . Set u= X^2 . factor out will get (u+1)(u-4)(4u-1)(4u^2+13u+4) =0
u can be only 4 and 1/4 . done
At 13:00 you could have tried to factor the equation - it simply comes to (4y-25)(4y+5) = 0. Saves a lot of maths.. Really interesting problem and solution nevertheless, thanks.
Does x^2 +1/x^2 = (x+1/x)^2 ? If I take (x + 1/x) ( x+1/x) I get x^2 + 1 + 1 + 1/x^2 = x^2 +2 + 1/x^2 which does not equal x^2 +1/x^2 ? AM I doing something wrong? IF not then how can x^5 + 1/x^5 = (x +1/5)^5
Bravo!!
je vous me chercher
Comment résolu
2ex(x)-3ex(y)=5
j'ai trouvé
x=3
y=1
maisl
ilya une solution
x=5
y=3
2×2×2×2×2=32
3×3×3=27
32-27=5
(S.V.P.)
Comment fait pour trouver la deuxième solution
Merci
Mes salutations
Amazing
But the complex solution are more real than real solution. The polynomials (with real coefficients) have all solutions in the complex numbers.
That's really true! 😁
I like your move on this one
Thank you!
Cross multiply, cancel the x (which can't be zero), and get this: 16 x^10 - 205 x^6 - 205 x^4 + 16 = 0. Now factor out x² + 1.
16(x² + 1)(x^8 - x^6 + x⁴ - x² + 1) - 205x⁴(x² + 1) = 0, and divide by (x² + 1) which can't be zero. 0 = 16(x^8 - x^6 + x⁴ - x² + 1) - 205x⁴.
Now divide by x⁴: 0 = 16(x⁴ - x² + 2 - 1/x² + 1/x⁴ - 1) - 205 = 16(x² + 1/x²)² - 16(x² + 1/x²) - 16 - 205. Solve this quadratic:
x² + 1/x² = (16 ± sqrt(256 + 4*16*221))/32 = ½ ± sqrt(4 + 221)/4 = ½ ± 15/4. x² + 1/x² is positive, so x² + 1/x² = 17/4. We get another quadratic whose solutions are reciprocals. x⁴ - 17x²/4 + 1 = 0. x² = (17/4 ± sqrt(289/16 - 4))/2 = 17/8 ± sqrt(289 - 4*16)/8 = 17/8 ± sqrt(225)/8 = (17 ± 15)/8 = 4 or 1/4. That means x = ±2 or ±½. Those are the only four real solutions.
Hello sir suggest me books for mathematics
Stop drawing complex roots away.
6complex sol. And +-2 and +-1/2 are real sol
Why not dividendo x*4_1 will be afactor_x*2_1)(x*2+1)/x*2+1 indenom which will disappear
Sure!
205/16=(x^10+1)/(x^4*(x^2+1))= u=x^2
(u^5+1)/(u^2(u+1))=(u^5+u^4-u^4-u^3+u^3+u^2-u^2-u+u+1)/(u^2(u+1))=
(u^4-u^3+u^2-u+1)/u^2=u^2+1/u^2-(u+1/u)+1=
(u+1/u)^2-(u+1/u)-1=205/16 T=u+1/u
t^2-t-221/16=0
t=17/4 (t deve essere positivo)
x^2+1/x^2=17/4
x^4-17/4x^2+1=0
x=+-sqrt((17+-15)/8)
so x=+-0.5 or x=+-2
ものすごいわかりやすい！
thanks!
It is incredible how the method to resolve specific polynomial equations of 4th degree can be used to resolve these kind of equations of higher degree.
Also, since we have found 4 real solution, the 6 ones left are all complex :P
I did this the old-fashioned way: I guessed x=2, and I was right. 17 minutes saved.
Well you can't just guess what the answer is.
I think cauchy can be used in this equation
Ngl, I saw the thumbnail, but I missed the part about real solutions. I ended up finding all 8 solutions (since 0, i, and -i all don’t work), and it was a piece of work
My method was like such:
Since we can obviously see that x can’t equal 0, the LHS can be divided by x/x. This leaves us with x^10 + 1 on top and x^6 + x^4 on the bottom. I was curious if integer solutions would work, so I tried x=2, and the top became 1025, or 5*205. I checked the bottom, and it matched up, so x=2 (and therefore x=-2 since it’s an even function now). From here, I substituted u=x^2, and I know the new equation has a root of u=4. Multiple denominators across and polynomial long division from here, and the remaining part we need to check for solutions is [u^4 + 64u^3 + 51u^2 - u - 4]
The new equation has an obvious root of -1 (which can be ignored, since this gives x values of i and -i, and these don’t work), so long division again. We now have [16u^3 + 48u^2 + 3u - 4] left to check for solutions. From here, I noticed that there will be one positive solution where u is less than one, and by fluke I found that u=(1/4) was a solution (I was checking “nice” to approximate what the root would be. I found this after realising 1/2 was way too big). Having this as a root of the expression means that (4x-1) is also a root, and long division happens here. Now there’s just a quadratic left, and so the rest is obvious. The quadratic only has negative u values as roots, so this can also be ignored (I wish I’d realised that before xD. I had to use a calculator to check them)
Probably not the most reliable method, as I totally got some of the roots by dumb luck, but at least I got to relearn how Synthetic Division works. Never would’ve thought of solving it via your method, and I certainly imagine that your way is more reliable
Nice work!
16u⁴-80u²-125 can be factorized as
(4u²-25)(4u²+5)
at first glance.
Amazing solution...Thank u very much..!!!
Most welcome!
interesting solution sir thank u
Welcome
Y equals 4 u squared is a much nicer substitution at the end. Leaves you with a much simplified calculation at the end.
This is how I relax
My goodness, I would never find the trick to solve the equation.
Excellent solution.very creative.
Thank you very much!
@@SyberMath OH COME ON NO ONE WOULD.EVER THINK OF THIS SOLUTION..I FACTORED OUT AN X FROM TOP AND BOTTOM NO ONE WOULD EVER THINK OF FSCTORINF OUT X TK THE FIFTH FROM THE BOTTOM..its random and infuriating..Why the fuck woukd anyone ever think of that? It shouldn't count
.
@@leif1075 man you have to mould your thinking like that if you are preparing for olympiads...and that comes with time.. It's not an obvious approach tho.
The answer is x is equal to 2.
y 因式分解真的會比較快
You deserve millions of subs and likes!!!!!!!!you make tooo good videos always have fun watching them!!
Thank you so much 😀💖
genial !!
good
square root of -1 is also a s o l u t i o n
So exquisit👌🏽👌🏽
Thank you 😊
Fantastic
Thank you so much 😀
Great learning algebra.
Thanks
Try a substitution x+x^(-2),
I mean x^2+x^(-2)
Beautiful solution
Thanks a lot
👍
Thank you sir
Welcome
Nice solution
Thank you!
Another great solution. Thank you!
You're welcome!
Is this type of subtitution common?
yes it is
Is this doable? 😁
Absolutely!
@@SyberMath Great problem! Thanks for the video. 😄
You're very welcome, sir!
SO COOL!
pro you are a pro!
Thank you!
@@SyberMath SO WELCOME YOU ARE!
Awesome.
Glad you think so!