Solving x^3+x-10=0 in Two Ways

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[j.r.]

Every time I see a demo of the so-called cubic formula, I become convinced that it's the hardest possible approach to these problems. Don't think we didn't notice how many steps you skipped over in the calculation! :) By far the easiest way to solve this particular equation (not necessarily every cubic) is to find the easy-to-spot integer root, then use long division, then solve the remaining quadratic. I do think it makes sense to always check first for simple integer roots: 0, 1, -1, 2, etc.

[minhdoantuan]

Let f(x)= x^3 + x - 10, we have f'(x) = 3x^2 + 1 > 0, so f(x) increases in its domain, so f(x) = 0 only has 1 real solution, bc y=0 is a constant.
When f(x) = 0, x^3 + x = 10. Easily checked that x = 2 satisfied, so x = 2 is the only real solution.

[stvp]

Now I want to see how those complicated radicals reduce to 2

[prasantbhattarai]

Let: f(x)= x^3+x-10 Here‚ The possible factor of -10 are +-1‚+-2‚+-5 and +-10
If x=2 then f(2)=0 then (x-2) Is a factor of it‚ By factor theorem and other factor you can get by dividing f(x) by (x-2).

[alnitaka]

The cubic formula not only is the hardest, but frequently gives incomprehensible results, such as this one giving a complicated expression for 2. It takes to tango.

[aguspueyrredon]

x³+x=10
x(x²+1)=10
Assuming we're dealing with natural numbers, you factorise 10 as either 1x10 or 2x5
easily noticeable that
x(x²+1)=2x5
Therefore X=2

[Alphabet]

it's possible to simplify the cube-root expression at 4:47 to obtain all 3 solutions at once! if you assume--as is reasonable to for problems like this--that the real solution is an integer (for those familiar with galois theory, it should be easy to convince yourself that if the real solution has a "simpler form" it should indeed be an integer via rational root theorem) then you can take the cube roots one by one. the first one should just be 1/3 times the cube root of 135+78*sqrt3, and treating 135+78*sqrt3 as the cube of a+b*sqrt3 for some integers a and b yields a couple diophantine equations that should be pretty easy to find a solution for. the other cube root is just the conjugate of this one, also easy to find as the solution works exactly the same way. ultimately the general solution should be (1+2/sqrt3)*w+(1-2/sqrt3)*w', where w is one of the 3 third roots of unity and w' is its complex conjugate (also a third root of unity). plugging in each third root of unity yields the solutions 2, -1+2i, -1-2i

[jarikosonen]

Great derivation for a kind of cubic formula...

[HoSza]

Educated guess method: first try with integer divisors of the constant term. Yes x=2 works, so divide both sides by x-2 and solve the second order equation you get. x2+2x+5=0. x=-1+-2i

[SuperYoonHo]

Wow! At first it looked so hard but then i thought... HEY... if x=2 then 8+2-10=0
so x=2 works! And then it worked out totally fined use and see that
x=2 and -1+or-2i

[angelmendez-rivera]

Using the integer roots theorem, if the polynomial x^3 + x – 10 does have an integer root, then it must be a divisor of –10. The set of divisors of 10 is {–10, –5, –2, –1, 1, 2, 5, 10}. It can be verified that, of these, 2 is a root.

(x^3 + x – 10)/(x – 2) = (x^3 – 2·x^2 + 2·x^2 – 4·x + 5·x – 10)/(x – 2) = x^2 + 2·x + 5 = (x + 1)^2 + 4 = (x + 1 + 2·i)·(x – 2·i), hence x^3 + x – 10 = (x – 2)·(x + 1 + 2·i)·(x + 1 – 2·i).

To use the cubic formula, x^3 + x – 10 = 0 is equivalent to 27·x^3 + 9·(3·x) – 270 = (3·x)^3 + 3·3·(3·x) – 270 = 0. Let y = 3·x, hence y^3 + 3·3·y – 270, hence y^3 = 3·(–3)·y + 270. Let y = u + v, thus u·v = –3 and u^3 + v^3 = 270, so u^3·v^3 = u^3·(270 – u^3) = 270·u^3 – u^6 = –27, hence u^6 – 270·u^3 – 27 = 0 = (u^3 – 27·5)^2 – 27^2·25 – 27 = (u^3 – 27·5)^2 – 27·2·337 = (u^3 – 27·5 + 3·sqrt(2·3·337))·(u^3 – 27·5 – 3·sqrt(2·3·337)). So y = ((1 + i·sqrt(3))/2)^m·cbrt(27·5 + 3·sqrt(2·3·337)) + ((1 + i·sqrt(3))/2)^(–m)·cbrt(27·5 – 3·sqrt(2·3·337)) with m = –1, 0, 1.

[ilana]

Dear teacher! Thanks for the great lessons! A big request: make the explanations a little slower and more detailed especially in this example. Thanks in advance!

[rubensramos]

Nice and easy: x =sqrt(Wq(200)/2) = 2, for q = 1/2. Wq(x) is the Lambert-Tsallis function.

[thegoldengood]

why is everyone using complicated methods i just went “is it one? oh it isn’t. is it two? oh it is”

[kanguru_]

x=2 is a solution by inspection. Factor that out then solve the quadratic.

[Dan-cwxu]

It's a good thing you weren't born in Greece. "Do you gamma what I gamma?" just doesn't have the same ring.

[SuperMtheory]

What?? The sum of the cube roots is 2? What algebraic manipulation gets that without a calculator?

[mcwulf]

I saw method 2 immediately and solved it very quickly then realised that x=2 was an obvious solution from the start.

[rakenzarnsworld]

x = 2
(x-2)(x^2+2x+5)
The other roots are imaginary number

[tontonbeber]

x=2 is obvious solution, so polynom factorize as (x-2)(x2+2x+5)
other solutions are (-2 +/- V(-16))/2 = -1 +/- 2j