Cubic Equations

[nycmathguy]

Set 1.3
Question 48
David Cohen

48. Solve each of the following equations for x. Hint: As in the text, begin by factoring out a common factor.

(a) x^3 -13x^2 + 42x = 0

(b) x^3 -6x^2 + x = 0

The above is from the David Cohen Precalculus textbook. The extra information below is not from the textbook.

Note: Do not solve 48 (a & b).

Factoring out the common factor for question 48 (a & b) is the logical thing to do as step one.

However, what if we were not able to factor out a common factor for a cubic equation?

Try these two:

(a) x^3 -13x^2 + 42 = 0

(b) x^3 -6x^2 +sqrt{2} = 0

Solve for x.

 

[MathLover1]

Solve for x:

(a)

x^3 -13x^2 + 42 = 0....cannot be factored

you can solve it using Newton- Raphson method
Newton's Method is based upon finding roots of a function f(x). To see how this applies to square or cube roots, suppose that y=√n for some fixed n. Well, then this y would be a root of the equation f(x)=x^2-n. Similarly, f(x)=x^3-n would provide us with a way to calculate the cube root of n.

or, you can solve it using The Cubic Formula
here is that formula:
https://math.vanderbilt.edu/schectex/courses/cubic/

as you can see, it is complicated to do
I suggest leave it for now and try those cubic equations that could be factored


(b) x^3 -6x^2 +sqrt(2) = 0....cannot be factored either

 

[nycmathguy]

Solve for x:

(a)

x^3 -13x^2 + 42 = 0....cannot be factored

you can solve it using Newton- Raphson method
Newton's Method is based upon finding roots of a function f(x). To see how this applies to square or cube roots, suppose that y=√n for some fixed n. Well, then this y would be a root of the equation f(x)=x^2-n. Similarly, f(x)=x^3-n would provide us with a way to calculate the cube root of n.

or, you can solve it using The Cubic Formula
here is that formula:
https://math.vanderbilt.edu/schectex/courses/cubic/

as you can see, it is complicated to do
I suggest leave it for now and try those cubic equations that could be factored


(b) x^3 -6x^2 +sqrt(2) = 0....cannot be factored either

It is beyond the scope of precalculus. I get it. Thanks again. I want to see more tutors here. I really miss freemathhelp.com. Very good tutors there.