A Cubic Equation with Radicals |
Вставка
- Опубліковано 5 лип 2024
- 🤩 Hello everyone, I'm very excited to bring you a new channel (aplusbi)
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡
/ @sybermathshorts
/ @aplusbi
Original post: / 1700257437230702912
⭐ Join this channel to get access to perks:→ bit.ly/3cBgfR1
My merch → teespring.com/stores/sybermat...
Follow me → / sybermath
Subscribe → ua-cam.com/users/SyberMath?sub...
⭐ Suggest → forms.gle/A5bGhTyZqYw937W58
If you need to post a picture of your solution or idea:
intent/tweet?text...
#polynomialequations #polynomials #algebra #cubicformula
via @UA-cam @Apple @Desmos @NotabilityApp @googledocs @canva
x^3-3x=10sqrt(2)
An Exponential Equation | 2^{3^{5^x}} = 3^{2^{5^x}}: • An Exponential Equatio...
A Surprising Exponential Equation: • A Surprising Exponenti...
PLAYLISTS 🎵 :
Number Theory Problems: • Number Theory Problems
Challenging Math Problems: • Challenging Math Problems
Trigonometry Problems: • Trigonometry Problems
Diophantine Equations and Systems: • Diophantine Equations ...
Calculus: • Calculus
Directly we guess that 2sqrt(2) is a solution right away then we just have to factorize
By intutition: x^3 - 3x = x(x^2-3) = 10sqrt2. My first guess is that if x is Asqrt2 for some integer A, then A*(2A^2-3) = 10. I can only factor 10 as 2*5 so I guess that A is 2 (since 2*5^2-3 is way too large and therefore A can't be 5). Checking 2sqrt2 for x completes the exercise.
Wow one of the best sum❤
Through a bit of guess and test, I found x=2sqrt(2) to be a solution. By polynomial long division we can factor the polynomial:
(x-2sqrt(2))(x^2+2sqrt(2)x+5) = 0
To find the other two solutions you can use the quadratic formula:
x = (-2sqrt(2) +- sqrt(8-20))/2
= - sqrt(2) +- isqrt(3)
That's exactly what I did!
Saw right away that x = 2 yields 2 and x = 3 yields 18 so 2 < x < 3 for at least one value of x. We are looking at 10root2 or about 14. Obvious guess is obvious. Factor it out and quadratic it up, x also equals (-root2) +/- root3*i.
It is impossible to find a formula to denest every nested radical involving a cube root. The best that can be achieved is to find a technique that is likely to work on radicals that are able to be denested. Here is a simple one that succeeds in many cases which are possible to denest:
(a + d * sqrt(s))^(1/3) = (1/2) * (3*d*x0 - a)^(1/3) * (1 + sqrt(s)/x0)
where x0 is a root of the cubic polynomial
d*x^3 - 3*a*x^2 + 3*d*s*x - a*s = 0
Obviously, this only denests the radical if you can find a rational root to the cubic polynomial. And that depends on whether any of the candidates from the rational root theorem actually satisfy the equation. So denesting the radical has been transformed to finding a rational root for a cubic polynomial.
The cube root mentioned in the video fails since there is no rational root.
Good point. I attempted it and got stuck
корень 2*sqrt(2) подбирается очень быстро. после этого остаётся квадратное уравнение.
u r helping me alot in my jee prep sir please upload some binomial theorem problems
ugh! 😜
Yeah, it was clear that x would have a sqrt(2) component. But why it was not considered that it could have a real component as well? So x=a*sqrt(2) +b. We could miss other solutions.
In our case, we have just one solution, as shown by the graph, so a=2 and b=0.
So, maybe we need to start with the graph and check how many solutions there could be.
Or try to find the other solutions by dividing the polynomial x^3-3x-10*sqrt(2) by x-2*sqrt(2) => (x-2sqrt(2))(x^2 +bx +c)=x^3-3x-10sqrt(2) => b=2sqrt(2), c=5. So, we are looking for the solutions of x^2+2sqrt(2)*x+5=0 and see that there are no real roots.
Good thinking! knowledge illusion
Guess and check x1 = 2√2
Very fun and easy problem! I'd love it if you could do more similar to this one.
I solved this by using rational root theorem, but adding a root(2) to the end of the possible rational roots (instead of checking ±1 ±2 ±5 ±10 i checked ±root(2) ±2root(2)....)
I ended up with x = 2root(2) as a solution and divided by it, getting a quadratic.
Solving by quadratic formula, i got 1 real solution and 2 complex solutions.
X = 2root(2)
X = -root(2) ± root(3)i
x=2√2
For what it's worth the method you used to find a and b goes back to the Mesopotamians.
Yes.
روعة
😂Cubica con D=7>0,1 soluzione reale e 2 complesse... X=sqrt3(5rad2-7)+sqrt3(5rad2+7)
Поищем корень среди делителей свободного члена, есть такой -2*2#1/2 разделим многочлен на х-2*2#1/2 и получим квадратное уравнение. Все.
Это работает только в случае уравнения с целыми коэффициентами. И вообще, сам подумай, у тебя есть иррациональное число и ты перебираешь его делители... Это как вообще, какие делители могут быть у иррационального числа?) Но в данном случае это просто совпадение.
Answer: 2.828