Question

How to solve cubic equation by cardon's method?Give an example.

Answer 1

The Cardano's formula (named after Girolamo Cardano 1501-1576), which is similar to the perfect-square method to quadratic equations, is a standard way to find a real root of a cubic equation like

We can then find the other two roots (real or complex) by polynomial division and the quadratic formula. The solution has two steps. We first "depress" the cubic equation and then solve the depressed equation.

To depress the cubic equation, we substitute

Then we get

Note that the above equation has been depressed in such a way that it is without the  term. Let's try this on an example.

Depress the following equation: Substituting  the equation is depressed to

The cubic polynomial

can be expressed in terms of  as

where  and  are real numbers, and  has a linear relationship with 

What is the value of 

Submit your answer

Solving the Depressed Equation

The next step is to solve the depressed equation of the form .

We first substitute

Then solve for

Find the real root of the cubic equation From the above example, we know the given equation is depressed into  Then we substituteFrom  we have  substituting which into  givesMultiplying  throughout by , we obtain the quadratic equation for  as follows:Solving for  and then , we haveIt can be shown that whether we take the positive or negative root of  we will get the same value for . We shall only consider the positive here.From  we havewhich impliesTherefore, the real root of the original cubic equation  isNote: We check that  is the only real root of of the cubic equation  because  can be factorized as Cardano's FormulaThe cubic polynomial  has solutionswhere  and in turn . 

After depressing the equation and making it monic by dividing by  we getNow consider the identity . If we make it match with the equation we getCube both sides of the second equation to get . Now, by Vieta's formula, the polynomial  will have roots  and , so let's solve it using the quadratic formula:Notice that the system of equations is symmetric in  and , so the order we choose doesn't matter, and the value of  will be the same. Sowhere  and  is any  primitive root of unity. We see that then we have 9 combinations for the value of , but only 3 of them work. By looking at the second equation, we see that  must be a multiple of 3, so  and our solutions areWe choose  and , soUndo the change  and we get our desired solutions. 

Find the roots of the following polynomial equation using Cardano's method:It is given thatCompute  and Compute  and Compute the roots:

Consider the following equatioN