A Cubic Equation with Radicals |
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x^3-3x=10sqrt(2)
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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.
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!
Wow one of the best sum❤
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
u r helping me alot in my jee prep sir please upload some binomial theorem problems
ugh! 😜
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.
корень 2*sqrt(2) подбирается очень быстро. после этого остаётся квадратное уравнение.
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
For what it's worth the method you used to find a and b goes back to the Mesopotamians.
x=2√2
Yes.
روعة
Answer: 2.828
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.
😂Cubica con D=7>0,1 soluzione reale e 2 complesse... X=sqrt3(5rad2-7)+sqrt3(5rad2+7)