I have done it much easily I let t = x^3 and i have a cubic equation and i noticed that \ / 2021 is a root of the equation so i factored by (t - \ / 2021) and got a quadratic equation that we ko how to solve, and got the values of t and I used the cubic root to get the values of x but i think i'm fortuate to find the root of the cubic equation quickly.
I have been "Binge-Solving" your videos lately and this question was the best of the lot ! Also I was able to solve it on my own ! And I used the exact same method which you did !
I tried to graph this on my phone's app. Oddly, 0.28119 was the only value graphed as far as I could see. But when I ventured further, the other 2 showed up as close roots. Maybe the slope is so drastic the app can't draw it.
At 14:05 you say: product of two primes is always the difference of two squares. Yes if the prime is not 2. And they do not have to be primes; just odd numbers. Because a•b = ((a+b)/2)^2 - ((a-b)/2)^2.
If you let u = x^3, the given equation becomes u^3 - 2022u + sqrt(2021) = 0. Because the cube of sqrt(2021) is 2021*sqrt(2021), u = sqrt(2021) is an obvious solution to u^3 - 2022u + sqrt(2021) = 0. If we use synthetic division to divide u^3 - 2022u + sqrt(2021) by u - sqrt(2021), our equation becomes (u - sqrt(2021))(u^2 + sqrt(2021)*u - 1) = 0. Using the quadratic formula on u^2 + sqrt(2021)*u - 1 = 0, we get two additional u-values: u = (-sqrt(2021) + 45)/2, along with u = (-sqrt(2021) - 45)/2. Taking the cube root of these three u-values yields the solutions presented in this video.
I began the same way but couldn't find the root intially by hit and trial then tried with another approach which was a bit lengthy , but this seems to be the best and most elegant one tnks
Take y=x^3 then define g(y)=y^3-(a^2+1)*y+a. Note that g(a)=0. Well y^3-a^2*y-y+a=y*(y^2-a^2)-(y-a)= (y-a)*(y*(y+a)-1)=(y-a)*(y^2+a*y-1) Then y=a or 2*y=-a+-√(a^2+4)=-a+-√(2021+4)=-a+-√(2025)=-a+-√(5^2*(81))=-a+-45.
Very neat solution. I do have a slight unease about it. The success of the strategy looks to me to be dependant on the expression under the first square root being factorizable into a perfect square. If this had not been the case all you would have got was another ugly expression. So was this factorization something that could have been seen from the start? I don't like taking a strategy which relies on hoping for the best.
Thank you! You mean the expression cube root([(-sqrt(2021)+45]/2)? The solution method would not be different even if we had a non-perfect square instead of 2025 under the radical if I understood your question correctly.
@@SyberMath It's not easy to be clear in describing these things. But the square root I had in mind was the one used in the expression for 'a'. That is when you used the quadratic formula to find 'a' in terms of 'x'.
Oh, I see. Thanks for the clarification. Yes, you're right. In order for this technique to work, we need a nice expression under the radical. In other words, the discriminant must be a perfect square.
*I don't like taking a strategy which relies on hoping for the best.* This is a little silly, though. Factorization, as a whole, is just a "let's hope for the best" method, where, if the coefficients are not the right type, then you just cannot factor the polynomial. So, what you're basically arguing here is that the only method we should ever rely on to solve these equations is Cardano's method.
I got the solutions you were looking for! However, since it's a nonic polynomial, shouldn't it have 9 solutions altogether? These three solutions are real, but wouldn't that mean the other six are complex?
Let y = x^3 and n^2 = 2021, so LHS = y^3 - (n^2+1)y + n = y(y^2 - n^2) - (y-n) = y(y+n)(y-n) - (y-n) = (y-n) (y^2+ny -1) = (Linear term 1 root) x (Quadratic term 2 roots) = 0 = RHS Therefore, this cubic has 3 solutions, y = n and y = [ -n +/- √(n^2+4) ] / 2 = [ +/-45 - n ] / 2, because Luckily, n^2 + 4 = 2021 + 4 = 2025 = 45^2 Each of these 3 real numbers (say y1,y2 and y3) have 1 real cube-root (say x1,x2 and x3) and a pair of complex conjugate roots (x1, x2, and x3 times complex conjugate cube-roots of unity). That includes all 9 (3 real and 6 complex conjugate) roots of the nonic equation as expected. *Simple* Right ?
If you set 2022 = n then you have y^3-ny+sqrt(n-1) =0 If you divide by (y-sqrt(n-1)) (Ruffini) you find that the division is exact with the result y^2-sqrt(n-1)y-1 Which can be solved to obtain the other two solutions.
Yes you could! Actually you could proceed as follows: set u=x^3 then u^3-2022u+sqrt(2021)=u^3-2021u-u+sqrt(2021)=u(u^2-2021)-(u-sqrt(2021))=0 and use difference of two squares
My solution (of course way faster), with ALL the solutions (not only the real ones): Let's call x^3=y. Our equation becomes: y^3-2022.y+sqrt(2021)=0. We immediately see that x1=sqrt(2021) is solution of the equation: (sqrt(2021))^3-2022.sqrt(2021)+sqrt(2021)=2021.sqrt(2021)-2022.sqrt(2021)+sqrt(2021)=0. So we can factorise by (y-sqrt(2021)). Our equation becomes: (y-sqrt(2021)).(y²+sqrt(2021).y-1)=0. Now let's solve the quadratic equation: Delta=2021+4=2025=45². So our solutions are x2=[-sqrt(2021)+45]/2 and x3=[-sqrt(2021)-45]/2. And now we solve x^3=y for x for y=x1,x2,x3. If we call x'1=(x1)^(1/3), x'2=(x2)^(1/3) and x'3=(x3)^(1/3), the nine solutions are: x'1, x'1.j, x'1.j², x'2, x'2.j, x'2.j², x'3, x'3.j and x'3.j² where j=exp(2.i.pi/3).
You don't need that, my friend, and with the cube root you don't have all the solutions (same issue than in the video). Here is my solution, way faster and with all the solutions: Let's call x^3=y. Our equation becomes: y^3-2022.y+sqrt(2021)=0. We immediately see that x1=sqrt(2021) is solution of the equation: (sqrt(2021))^3-2022.sqrt(2021)+sqrt(2021)=2021.sqrt(2021)-2022.sqrt(2021)+sqrt(2021)=0. So we can factorise by (y-sqrt(2021)). Our equation becomes: (y-sqrt(2021)).(y²+sqrt(2021).y-1)=0. Now let's solve the quadratic equation: Delta=2021+4=2025=45². So our solutions are x2=[-sqrt(2021)+45]/2 and x3=[-sqrt(2021)-45]/2. And now we solve x^3=y for x for y=x1,x2,x3. If we call x'1=(x1)^(1/3), x'2=(x2)^(1/3) and x'3=(x3)^(1/3), the nine solutions are: x'1, x'1.j, x'1.j², x'2, x'2.j, x'2.j², x'3, x'3.j and x'3.j² where j=exp(2.i.pi/3).
You made a mistake so the problem turned into big. In the third step, x^9-a^2.x^3-x^3+a=0, this equation was factorible and we easily get the solutions.
But don't forget (like our friend did) that when you solve x^3=y for x you have three solutions: y^(1/3), y^(1/3).j and y^(1/3).j² (where j=exp(2.i.pi/3).
put y = x*x*x and n = 2021 to get (( n+1) - y*y)y = √n y = √n being a solution to this one, this can be rewritten as (y-√n)(y*y +y√n -1) = 0 or y = √n, (-√n+ √(n+4))/2, (-√n- √(n+4))/2 x = y, omega*y, omega^2*y, are the final answers
I have done it much easily
I let t = x^3
and i have a cubic equation and i noticed that \ / 2021 is a root of the equation so i factored by (t - \ / 2021) and got a quadratic equation that we ko how to solve, and got the values of t and I used the cubic root to get the values of x
but i think i'm fortuate to find the root of the cubic equation quickly.
Same
this popped into my head during 2nd minute of video.
wondered why strategy in video had to be so complicated
This technique is quite powerful! Ive now seen many questions of this type
A little easier I think if you substitute for y=x^3 at the beginning. Easier to see that y^3 - (a^2 + 1)y + a = 0 is (y - a)(y^2 + ay - 1) = 0 etc
yess
I have been "Binge-Solving" your videos lately and this question was the best of the lot !
Also I was able to solve it on my own ! And I used the exact same method which you did !
Great job!
@@SyberMath isnt there another way tomsolve i dont see how or ehy anyone would solve it that way..why not u subidtituon with x cubed
You mean substitute something for x cubed at the beginning?
@@SyberMath yea u equals x cubed
srijan bhowmick - maybe u r bengali?
At 3:16 The equation factorises to (x^3-a)(x^6+ax^3-1) = 0, so you are finished. Very clever substitution though.
Thanks!
I tried to graph this on my phone's app. Oddly, 0.28119 was the only value graphed as far as I could see. But when I ventured further, the other 2 showed up as close roots. Maybe the slope is so drastic the app can't draw it.
Probably
At 14:05 you say: product of two primes is always the difference of two squares. Yes if the prime is not 2. And they do not have to be primes; just odd numbers. Because a•b = ((a+b)/2)^2 - ((a-b)/2)^2.
Is it just a coincidence that the expression in the radical can be factored (6:33), or would this occur more often for equations like this?
When are you going to solve a comic equation? I can't wait.
That's a good one! We should also be doing atomic, conic, bionic, and polyphonic ones! 😂😂😂
@@SyberMath How about "pathetic"? 😂😂😂
@@SyberMath WHAT DOES THAT MEAN?
@@SyberMath THANKS!!!!!!!!
If you let u = x^3, the given equation becomes u^3 - 2022u + sqrt(2021) = 0. Because the cube of sqrt(2021) is 2021*sqrt(2021), u = sqrt(2021) is an obvious solution to u^3 - 2022u + sqrt(2021) = 0. If we use synthetic division to divide u^3 - 2022u + sqrt(2021) by u - sqrt(2021), our equation becomes (u - sqrt(2021))(u^2 + sqrt(2021)*u - 1) = 0. Using the quadratic formula on u^2 + sqrt(2021)*u - 1 = 0, we get two additional u-values: u = (-sqrt(2021) + 45)/2, along with u = (-sqrt(2021) - 45)/2. Taking the cube root of these three u-values yields the solutions presented in this video.
I began the same way but couldn't find the root intially by hit and trial then tried with another approach which was a bit lengthy , but this seems to be the best and most elegant one tnks
is it necessary to say continuously "ok cool"?
Whome this exercise is suited to?
I don't get why the other 6 solutions are complex
that was amazing!!!
I have really never thought about inverting parameter and variable. I'm astonished.
Very nice problem .Sooo happy
Thank you so much 😀
Take y=x^3 then define g(y)=y^3-(a^2+1)*y+a. Note that g(a)=0.
Well
y^3-a^2*y-y+a=y*(y^2-a^2)-(y-a)= (y-a)*(y*(y+a)-1)=(y-a)*(y^2+a*y-1) Then y=a or 2*y=-a+-√(a^2+4)=-a+-√(2021+4)=-a+-√(2025)=-a+-√(5^2*(81))=-a+-45.
Good!
Yeah~! I solved this exactly same way! Your other videos have inspired me!
Nice work!
x^9-a^2x^3-x^3+a = x^3(x^6-a^2)-(x^3-a). Now we can factorize first term using A^2-B^2 and take common.
Nice!
Absolutely , so easy ,Hemant Ji.Good approach
Very neat solution. I do have a slight unease about it. The success of the strategy looks to me to be dependant on the expression under the first square root being factorizable into a perfect square. If this had not been the case all you would have got was another ugly expression. So was this factorization something that could have been seen from the start? I don't like taking a strategy which relies on hoping for the best.
Thank you! You mean the expression cube root([(-sqrt(2021)+45]/2)? The solution method would not be different even if we had a non-perfect square instead of 2025 under the radical if I understood your question correctly.
@@SyberMath It's not easy to be clear in describing these things. But the square root I had in mind was the one used in the expression for 'a'. That is when you used the quadratic formula to find 'a' in terms of 'x'.
Oh, I see. Thanks for the clarification. Yes, you're right. In order for this technique to work, we need a nice expression under the radical. In other words, the discriminant must be a perfect square.
*I don't like taking a strategy which relies on hoping for the best.*
This is a little silly, though. Factorization, as a whole, is just a "let's hope for the best" method, where, if the coefficients are not the right type, then you just cannot factor the polynomial. So, what you're basically arguing here is that the only method we should ever rely on to solve these equations is Cardano's method.
I got the solutions you were looking for! However, since it's a nonic polynomial, shouldn't it have 9 solutions altogether? These three solutions are real, but wouldn't that mean the other six are complex?
yes
This is a very cool problem!
I totally agree. I'm glad you like it, too!
Nice approach .
Glad you think so!
Great video
Glad you enjoyed it
I enjoy watching your video
And thank u for teach us a new way of thinking
It's my pleasure!
Very very nice problem
Thanks!
3 real and 6 complex solutions, nice:
*Solve[x^9 - 2022 x^3 + Sqrt[2021] == 0] // N*
Let y = x^3 and n^2 = 2021, so LHS = y^3 - (n^2+1)y + n = y(y^2 - n^2) - (y-n) = y(y+n)(y-n) - (y-n) = (y-n) (y^2+ny -1) = (Linear term 1 root) x (Quadratic term 2 roots) = 0 = RHS
Therefore, this cubic has 3 solutions, y = n and y = [ -n +/- √(n^2+4) ] / 2 = [ +/-45 - n ] / 2, because Luckily, n^2 + 4 = 2021 + 4 = 2025 = 45^2
Each of these 3 real numbers (say y1,y2 and y3) have 1 real cube-root (say x1,x2 and x3) and a pair of complex conjugate roots (x1, x2, and x3 times complex conjugate cube-roots of unity).
That includes all 9 (3 real and 6 complex conjugate) roots of the nonic equation as expected. *Simple* Right ?
If you set 2022 = n then you have
y^3-ny+sqrt(n-1) =0
If you divide by (y-sqrt(n-1)) (Ruffini) you find that the division is exact with the result
y^2-sqrt(n-1)y-1
Which can be solved to obtain the other two solutions.
Nice!
Why couldn't you just take x cube as y and solve for y, by factorizing the cubic equation through intuition!?
This way is round about, yet interesting
Yes you could! Actually you could proceed as follows: set u=x^3 then u^3-2022u+sqrt(2021)=u^3-2021u-u+sqrt(2021)=u(u^2-2021)-(u-sqrt(2021))=0 and use difference of two squares
I just used the cubic formula after replacing x^3=u, and as it's a depressed cubic it's much easier to solve.
Namaste🙏 sirji💐.
Namaste! 🙏
Great
Thank you!
My solution (of course way faster), with ALL the solutions (not only the real ones):
Let's call x^3=y. Our equation becomes:
y^3-2022.y+sqrt(2021)=0.
We immediately see that x1=sqrt(2021) is solution of the equation:
(sqrt(2021))^3-2022.sqrt(2021)+sqrt(2021)=2021.sqrt(2021)-2022.sqrt(2021)+sqrt(2021)=0.
So we can factorise by (y-sqrt(2021)). Our equation becomes:
(y-sqrt(2021)).(y²+sqrt(2021).y-1)=0.
Now let's solve the quadratic equation: Delta=2021+4=2025=45².
So our solutions are x2=[-sqrt(2021)+45]/2 and x3=[-sqrt(2021)-45]/2.
And now we solve x^3=y for x for y=x1,x2,x3. If we call x'1=(x1)^(1/3), x'2=(x2)^(1/3) and x'3=(x3)^(1/3), the nine solutions are:
x'1, x'1.j, x'1.j², x'2, x'2.j, x'2.j², x'3, x'3.j and x'3.j² where j=exp(2.i.pi/3).
So cool.tx
You're welcome
I am Indian students
Very tremendous teacher's
Thank you!
Just substitute x^3= u solve for u with cardano formula and then cube root u and you get the answers.
You don't need that, my friend, and with the cube root you don't have all the solutions (same issue than in the video). Here is my solution, way faster and with all the solutions:
Let's call x^3=y. Our equation becomes:
y^3-2022.y+sqrt(2021)=0.
We immediately see that x1=sqrt(2021) is solution of the equation:
(sqrt(2021))^3-2022.sqrt(2021)+sqrt(2021)=2021.sqrt(2021)-2022.sqrt(2021)+sqrt(2021)=0.
So we can factorise by (y-sqrt(2021)). Our equation becomes:
(y-sqrt(2021)).(y²+sqrt(2021).y-1)=0.
Now let's solve the quadratic equation: Delta=2021+4=2025=45².
So our solutions are x2=[-sqrt(2021)+45]/2 and x3=[-sqrt(2021)-45]/2.
And now we solve x^3=y for x for y=x1,x2,x3. If we call x'1=(x1)^(1/3), x'2=(x2)^(1/3) and x'3=(x3)^(1/3), the nine solutions are:
x'1, x'1.j, x'1.j², x'2, x'2.j, x'2.j², x'3, x'3.j and x'3.j² where j=exp(2.i.pi/3).
@@italixgaming915 omg I did not even think a bit about that.
You made a mistake so the problem turned into big. In the third step, x^9-a^2.x^3-x^3+a=0, this equation was factorible and we easily get the solutions.
let x³=y,through observing,one of the solution is √2021，so,equation divided by y-√2021,we got( y²+√2021y-y)(y-√2021),the answer is same as yours.
But don't forget (like our friend did) that when you solve x^3=y for x you have three solutions: y^(1/3), y^(1/3).j and y^(1/3).j² (where j=exp(2.i.pi/3).
Come on, lets see the work for all 9
A monic nonic!
😁
7:21 😂
Murphy's Rules!
lol
put y = x*x*x and n = 2021
to get (( n+1) - y*y)y = √n
y = √n being a solution to this one, this can be rewritten as
(y-√n)(y*y +y√n -1) = 0
or y = √n, (-√n+ √(n+4))/2,
(-√n- √(n+4))/2
x = y, omega*y, omega^2*y,
are the final answers
A few moments later
Even though it looks nonic, in fact it's cubic
Right
@@SyberMath 😉
3|2022
🔥
I prefer solving cubic
That makes sense
When the subject is confusing yet the teacher keeps talking annoyingly.
That's me! 😂
Круто
Спасибо
0.28119054
This way no good