# Thread: Problem #4 - Solving a cubic equation

1. ## Problem #4 - Solving a cubic equation

Solve for x:
$x^3 + 4x^2 - 3 = 0$
without using the rational root theorem.

Hint: There is one integer solution.

There is a way to factor this but if you actually want to slog through it using Cardano's Method I'll give you a hero cookie.

-Dan

2. ## Re: Problem #4 - Solving a cubic equation

Hello, topsquark!

$\text{Solve for }x\text{ without using the Rational Root Theorem: }\;x^3 + 4x^2 - 3 \:=\: 0$

Spoiler:

$\begin{array}{cccc}4x^2 = x^2 + 3x^2: & x^3 + {\color{blue}x^2 + 3x^2} - 3 \:=\:0 \\ \\ \text{Add/subtract }3x: & x^3 + x^2 + 3x^2 {\color{blue}\,+\,3x - 3x} - 3 \:=\:0 \\ \\ \text{Factor:} & x^2(x+1)+3x(x+1)-3(x+1) \:=\:0 \\ \\ \text{Factor:} & (x+1)(x^2+3x-3) \:=\:0 \end{array}$

$\begin{array}{cccccc}\text{Therefore:} &x+1 \:=\:0 & \Rightarrow & x \:=\:\text{-}1 \\ & x^2+3x-3\:=\:0 & \Rightarrow & x \:=\:\dfrac{\text{-}3\pm\sqrt{21}}{2} \end{array}$

3. ## Re: Problem #4 - Solving a cubic equation

Did you not understand topsquark's post? That is NOT Cardano's method so: NO cookie for you!

However, topsquark, Cardano's method solve the "reduced cubic" of the form $x^3+ mx= n$. Of course, any cubic can be put in that form but did you mean $x^3+ 4x- 3= 0$?

4. ## Re: Problem #4 - Solving a cubic equation

Originally Posted by HallsofIvy
Did you not understand topsquark's post? That is NOT Cardano's method so: NO cookie for you!

However, topsquark, Cardano's method solve the "reduced cubic" of the form $x^3+ mx= n$. Of course, any cubic can be put in that form but did you mean $x^3+ 4x- 3= 0$?
The problem is correct as written. Of course, as you say, you can change the form of the equation to put it into x^3 + mx + q by a translation of coordinates and in that form it may be "attacked" ala Cardano. I was hoping that no one would use it, but instead apply counter-terms to the problem.

Am I missing something? Soroban's solution is correct and to the best of my knowledge doesn't use Cardano's method?

-Dan

5. ## Re: Problem #4 - Solving a cubic equation

Spoiler:
here we can use numerical method
find interval using f(a)f(b)<0
so for the above eqn we get our interval as
(-1.5,0)
now we chose c=-1.5/2=-0.75
now we apply newton raphsons method
x(n+1)=x(n)-f(x)/f'(x)
here x1=-0.75-f(-0.75)/f'(-0.75)
x1=-1.02
do more iteration for more accurate value
our first iteration is giving soln closer to integer soln -1
so the integer soln of this eqn is -1

6. ## Re: Problem #4 - Solving a cubic equation

Originally Posted by prasum
Spoiler:
here we can use numerical method
find interval using f(a)f(b)<0
so for the above eqn we get our interval as
(-1.5,0)
now we chose c=-1.5/2=-0.75
now we apply newton raphsons method
x(n+1)=x(n)-f(x)/f'(x)
here x1=-0.75-f(-0.75)/f'(-0.75)
x1=-1.02
do more iteration for more accurate value
our first iteration is giving soln closer to integer soln -1
so the integer soln of this eqn is -1
It does the job, so good work. On the other hand don't you need to show convergence for this method?

-Dan

7. ## Re: Problem #4 - Solving a cubic equation

yeah we can check check the function for monotonicity
f'(x) should not be equal to 0 in (a,b) and f''(x)>=0 or <=0 for all x in (a,b) (for concavity)
at endpoints a,b mod(f(a)/f'(a))<b-a and mod(f(b)/f'(b))<b-a

i think all these conditions are holding for above

8. ## Re: Problem #4 - Solving a cubic equation

Looks like a good batch of solutions this time around. Keep up the good work!

-Dan