Solving Cubic Equations using Cardano's Rule

How to solve a cubic equation, using the classic version of Cardano's rule.
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[@musicsubicandcebu1774]

The best, fastest, clearest and most insightful explanation I've found - thankyou

[@Rokca719]

Just heard the professor explain this in class. I am so happy I find this video. Professor! You should share this with everyone via email :)

[@lupitabenavidez1766]

Thank you so much for this explanation. I reviewed my textbook so many times before watching this video. I wish I hadn't wasted so much time and have watched your video first. A simple explanation to an otherwise tedious problem.

[@jasonstemarie6878]

Thank you for demonstrating how to solve the depressed cubic in this way. It was interesting to see this approach with use of classic concepts such as adding and subtracting equations versus being given the purely algebraic version of Cardano's given Rule/Formula itself and solving by simply plugging in values.

[@scottleung9587]

Cool problem, Jeff! However, you could've simplified sqrt(108) to 6*sqrt(3).

[@donsena2013]

Does a cubic equation normally have three possible values of the cubed variable ?

[@rottershier6775]

Best explanation, thanks!

[@inoobbeyt1031]

Awesome man genius love it

[@jadekan72]

Two thing her: 1.at 3:10, why dose U cube plus V cube only equal the square root of 432? should it be also equal the negative square root of 432? 2. About the square root of 432, isn't it bedre with 12 times the square root 3 than 4 times the square root 108?

[@shandyverdyo7688]

So, since cubic has 3 roots. Then how to find the other 2 solutions using that cardano formula? Or is that formula only apply for 1 solution only?

[@gamerdio2503]

Multiply your answer by the cube roots of unity.

[@calvinjackson8110]

It's a pity this man is getting credit for something that doesn't belong to him at all. It's not Cardano's formula at all. From what I have read, he only published it. The one who discovered the method is the one who deserves the credit.

[@tristanseow2970]

So is it possible to solve the final expression? And could you make a tutorial on it? Thanks in advance

[@tristanseow2970]

I have been trying out different ways to solve the radical but mostly ended up returning to the original depressed cubic

[@JinxDojo]

If I'm reading your statement correctly: I've had the same curiosity for many years and have yet to find a solution. It "feels like" there should be a way of clearly identifying rational roots from the results of Cardano's formula. In the quadratic equation (or completing the square), integer/rational solutions will reveal themselves through root extraction. However, in the cubic results, there are cases in which it seems as though we have the sum of two (ostensibly) "irrational" numbers (the cube roots) somehow equivalent to a rational/integer number. (This alone seems noteworthy to me--even outside of solving cubics / algebra.) I recall trying all sorts of things years ago with cubic roots of unity, complex numbers, and even root-extraction algorithms to extract (or combine) the cubic roots in some manner, but everything I've read on the subject seems to be focused on combining/extracting complex roots, and makes no mention on how to show particular sums of two cubic roots to be rational/integer. (You can put such solutions into a calculator and they will clearly "complement" each other to produce the rational/integer number.) Of course, the rational root test is always available--and, indeed, I'm left with the impression that, despite it feeling rather "backward" in process, this is actually "the solution" to combining such radicals rather than the other way around--and therefore perhaps even the only route to finding rational roots without a calculator (at least non-obvious ones). Perhaps there is just some idealism in me that fervently desires the cubic results to behave like the quadratic, and reduce in scope when applicable (i.e. from real to rational), but there is actually no mathematical reason why it must? Do share if you (or anyone) ever finds a method of handling such cubic results (or a proof that such a thing can't be done and necessarily reverts back to the rational root test.) Thanks!

[@briana5708]

I agree Jinx. I don’t find the cubic formula very satisfying because it seems that it always leads to more computing. And I did the same here to try to simplify the final answer but it leads back to the depressed cubic

[@julicz666]

i can finally sleep in peace

[@rottershier6775]

Why do you add 4u^3v^3 there? Whats the purpose?

[@rabonssons]

to complete the square. this way the left side can be simplified

[@sgjbslover]

shouldnt 3uv=-6?

[@michaeltuchman9656]

Excellent presentation.

[@esamh1640]

Thank you a lot, Please: can you tell me a names of some references contain some exercise for cubic equation ? thank you for your work.

[@astrophysicscosmology7935]

Thanks sit... So much thanks.. 😊😊😊

[@Hamad-Saif]

I don't know where the rests of solution is! It's supposed that there are three values for "X" I

[@JeffSuzukiRandomProfessor]

Once you know one solution, you can divide by the factor corresponding to that solution; this reduces the equation to a quadratic, which can be solved normally.

[@jean-philippesuter2550]

@@JeffSuzukiRandomProfessor Does that give the same extra two solutions as the method described by @gamerdio2503, namely multiplying the original solution by the cube roots of 1?

[@julicz666]

but i wonder why the heck people put these in a 30 minute exam with another 4 questions

[@Mucurex]

And the Complex answers? Where are them?

[@JeffSuzukiRandomProfessor]

You can find them one of two ways. First, you can divide: If x = a makes a polynomial equal to zero, then x - a is a root. (kzbin.info/www/bejne/e2ncpqSXZsqribs). However, that's a challenge if the roots are complicated experssions. The alternative is to use the fact that the product of the three roots is the constant, while the sum is the coefficient of the quadratic term. In this case, since you know one root, that means you know the product of the two other roots, and the sum of the two other roots; this allows you to find the roots themselves using the quadratic formula.

[@gemeni0]

Хорошо, что автор понимает, что объясняет. Плохо то, что объяснять не умеет. На вроде шутки самосмейки: сам шучу - сам смеюсь.

[@IronMaidenEE]

Actually x=2 satisfies the equation I think

[@JeffSuzukiRandomProfessor]

In fact, x = 2 is the only real solution, which raises an important problem: the rather horrific expression at the end MUST be equal to 2...so how do you reduce it? This played a role in the subsequent history of mathematics.

[@IronMaidenEE]

Jeff Suzuki is there a way to calculate that horrific expression in the end so that we get x = 2?

[@JeffSuzukiRandomProfessor]

There is...but it's not pretty. What you'd do is assume it can be reduced to an expression of the form a + b R, where a, b are rational and R is some radical expression (here, probably sqrt(108)). Then some really awful algebra ensues. (If you've ever solved equations involving square roots, the basic idea is the same, though there's a twist that makes it a lot more complicated)

[@willjohnston2959]

@@IronMaidenEE if 10 + √108 were actually the cube of something, we could simplify things. Since 10 + √108 = 10 + 6√3 you may perhaps think it could be an expression of the form (a + b√3)^3. Expand that to get a^3 + 3a^2* b√3 + 3ab^2 * 3 + b^3 * 3√3. Organize the terms: (a^3 + 9ab^2) + (3a^2 * b + 3b^3) √3 Is there a way to get that to match 10 + 6√3? We'd need a^3 + 9ab^2 = 10, so (a)(a^2 + 9b^2) = 10. We'd also need 3a^2b + 3b^3 = 6, so (3b)(a^2 + b^2) = 6. In this case a little experimentation shows that integers a = 1 and b = 1 work. So 10 + √108 actually is (1 + √3)^3. u = cbrt(10 + √108) = cbrt((1 + 1√3)^3) = 1 + √3 Similar reasoning for v^3 = √108 - 10 = 6√3 - 10 = (c√3 + d)^3 implies c^3 * 3√3 + 9c^2 * d + 3c√3 * d^2 + d^3, which is (3c^3 + 3cd^2)√3 + (9c^2 d + d^3) = 6√3 - 10 and that yields c = 1 and d = -1, so (√3 - 1)^3 is another name for sqrt(108) - 10. v = -1 + √3 is one cube root. Thus one answer is x = u - v = 1 + √3 - (-1 + √3) = 1 + √3 + 1 - √3 = 2. That means the complicated expression at end of this video can have the value 2. There are two other complex answers for u and for v found by multiplying each by complex values of cube root of 1. Then suitably adding pairs gives the two complex solutions. Or you can just use (x - 2) as a factor for x^3 + 6x -20 = 0 and after finding the quadratic factor use quadratic formula. Those other two values will be complex numbers and conjugates. (x^3 + 6x - 20) ÷ (x - 2) = x^2 + 2x + 10 Use quadratic formula. x = -2/2 +/- sqrt(2^2 - 4 * 1 * 10)/2 x = -1 +/- sqrt(4 - 40)/2 x = -1 +/- sqrt(-36)/2 x = -1 +/- 6i/2 x = -1 + 3i and x = -1 - 3i are the other two solutions, which you can verify in the original cubic if you know how to multiply complex numbers. Final answer: 2, -1 + 3i, -1 - 3i are the three roots of x^3 + 6x = 20.

[@ernestschoenmakers8181]

@@willjohnston2959 In fact i found the 2 by way of 1+sqrt(3)+1-sqrt(3) after plugging in the p and q where p=6 and q=-20.

[@ernestschoenmakers8181]

Can somebody solve this equation for me: x^3+x-1=0, please!!

[@willjohnston2959]

Following exactly the method of this video. x^3 + x - 1 = 0 x^3 + 1x = 1 where p = 1 (the coefficient of x term on left) and q = 1 (the constant term on right). Solution will be x = u - v provided that first criterion 3uv = p = 1 and second criterion u^3 - v^3 = q = 1 are satisfied. We will find u and v but do so by first finding u^3 + v^3. Start by squaring the second criterion. u^3 - v^3 = 1 (u^3 - v^3)^2 = 1^2 (u^3)^2 - 2(u^3)(v^3) + (v^3)^2 = 1 Add 4(u^3)(v^3) to both sides. (u^3)^2 + 2(u^3)(v^3) + (v^3)^2 = 1 + 4(u^3)(v^3) Simplify left side as a square. Simplify right side using first criterion that 3uv = 1 so uv = 1/3 and (u^3)(v^3) = (1/3)^3 = 1/27. (u^3 + v^3)^2 = 1 + 4 * (1/27) Take square root of both sides. u^3 + v^3 = √(1 + 4/27) u^3 + v^3 = √(31/27) Now combine that equation in a system with the first criterion. u^3 + v^3 = √(31/27) u^3 - v^3 = 1 Add the equations: 2u^3 = √(31/27) + 1 u^3 = √(31/27)/2 + 1/2 Subtract the equations. 2v^3 = √(31/27) - 1 v^3 = √(31/27)/2 - 1/2 Now express u as cube root of u^3 and v as cube root of v^3 and say x is the difference u - v x = ³√[√(31/27)/2 + 1/2] - ³√[√(31/27)/2 - 1/2] That is the final answer. Using a calculator ... x = u - v ≈ 1.0117801 - 0.3294523 ≈ 0.6823278 If you check that you will see it satisfies the original equation. (0.6823278)^3 + 0.6823278 - 1 = 0

[@ernestschoenmakers8181]

@@willjohnston2959Thanks for the confirmation, i saw that this equation has 1 real and 2 conjugated complex roots. I plugged in p=1 and q=-1 from the general depressed equation x^3+px+q in Cardano's formula and i got 0.68233 as the real root.

[@shandyverdyo7688]

So, since cubic has 3 roots. Then how to find the other 2 solutions using that cardano formula? Or is that formula only apply for 1 solution only?

[@MahmudinMahmudin]

You can do it with horner

[@shandyverdyo7688]

@@MahmudinMahmudin no, i mean by using cardano only. Actually there's a formula for general cardano's formula for finding the whole 3 roots (i forgot the site). But just don't know how to apply it.

[@MahmudinMahmudin]

@@shandyverdyo7688 yes. There is cubic formula for cubic equation with coeficient of x^2 is 0. I just started learning this.

[@esamh1640]

Thank you a lot, Please: can you tell me a names of some references contain some exercise for cubic equation ? thank you for your work.