Showing all solutions to a cubic equation

[Mr Davis 97]

I am solving the following problem: Find all real x such that $x^3 + 3x^2 + 3x = 1$

I complete the cube by adding 1 to both sides, and get that $(x + 1)^3 = 2$ then $x = 2^{1/3} - 1$

What I'm asking is how can I be sure that I have found all real solutions? What if there are other solutions?

 

[member 587159]

Do you know a formula to find the factors of x^3 - a^3 ?

 

[Mr Davis 97]

Ah, I see. If we factor it this way then we can find the other two complex conjugate roots, which means that with one real root and two complex roots, we have found all of the solutions since it is a cubic equation. However, finding the first root the way I did, how can I be sure that that is the only real solution, and that the other two are complex? Is the only way to show this through factoring the difference of the cubes like you said?

 

[mfb]

(x+1)³ in the real numbers is monotonous, the equation has to have exactly one real solution.

If the formula is more complex, you need more powerful tools to look for solutions.

 

[member 587159]

Ah, I see. If we factor it this way then we can find the other two complex conjugate roots, which means that with one real root and two complex roots, we have found all of the solutions since it is a cubic equation. However, finding the first root the way I did, how can I be sure that that is the only real solution, and that the other two are complex? Is the only way to show this through factoring the difference of the cubes like you said?

Here you can apply x^3 -a^3 = (x-a)(x^2 + ax + a^2). You immediately see that this only has one real solution because x^2 + ax + a^2 has a negative discriminant -3a^2, so there is only one real solution.

 

[mathman]

Let a be the known solution. You can use synthetic division (original cubic divided by x-a) to get a quadratic for the other 2 solutions.