Solving Cubic Equation: Positive Root Exists

  • Thread starter littleHilbert
  • Start date
  • Tags
    Cubic Roots
In summary, the conversation discusses how to show that for negative c, the equation x^3+ax^2+bx+c=0 has at least one positive root. The first approach involves considering the equivalent form of the equation for large |x| and showing that there is at least one real root. The second approach involves using the properties of cubic polynomials and comparing coefficients to show that at least one factor must be positive for c to be negative. The conversation also mentions the possibility of using calculus to solve the problem.
  • #1
littleHilbert
56
0

Homework Statement



Show that for negative c (a,b,c - real) equation [itex]x^3+ax^2+bx+c=0[/itex] has at least one positive root.
2. The attempt at a solution

Considering the equivalent form of the equation above for large |x|:
[itex]x^3(1+\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x^3})=0[/itex] we can conclude that there exists at least one real root, because the function changes signes if x does the same.

Let r,s,t be some roots of this equation. At least one of them is real. The cubic polynomial on LHS can be written in the form:
[itex](x-r)(x-s)(x-t)[/itex]. Thus the equation admits the following form: [itex]x^3-(r+s+t)x^2+(rs+st+rt)x-rst=0[/itex]. Comparing yields: -(r+s+t)=a, rs+st+rt=b and -rst=c. Now if for example r is real and so are s and t, then in order for c to be negative the product rst must be positive. This in turn is true only if at least one the factors is positive.

Is it correct?

Now the task says nothing about whether all the roots are assumed to be real or not. In case only one root is real, for example r, is it correct to say that the product st is real?

Is there another way to solve the problem?
 
Physics news on Phys.org
  • #2
littleHilbert said:
Now the task says nothing about whether all the roots are assumed to be real or not. In case only one root is real, for example r, is it correct to say that the product st is real?

If all are not real, then consider -rst=c. Supposing r is real, as above, then s and t must necessarily be complex conjugates of each other (otherwise the product rst would be complex, which is not allowed since c is real) , and so st>0, which gives r>0
 
  • #3
There's a good approach to this problem using calculus.
 
  • #4
littleHilbert said:

Is there another way to solve the problem?


You were getting close when you started thinking about changes of sign of the function. A polynomial is a continuous function, so if you can show it changes sign in some interval [0,X] for some (large) value of X, it has a root in that interval.
 
  • #5
Yes, he understood that. That's why he asked "is this another way to solve the problem".
 
  • #6
Mean Value Theorem.
 

1. What is a cubic equation?

A cubic equation is a polynomial equation of the form ax^3 + bx^2 + cx + d = 0, where a, b, c, and d are constants and x is the variable. It can also be written as x^3 + px^2 + qx + r = 0.

2. How do you know if a cubic equation has a positive root?

A cubic equation has a positive root if the constant term (d or r) is positive and the leading coefficient (a or 1) is negative. This is because a positive number multiplied by a negative number will result in a negative number, and a negative number multiplied by a negative number will result in a positive number.

3. What is the formula for solving a cubic equation with a positive root?

The formula for solving a cubic equation with a positive root is x = (sqrt((-q/2)^2 + (p/3)^3) - (q/2))^(1/3) - (p/3) / 3^(1/3). This formula is known as the Cardano formula and can be used to find the positive root of a cubic equation with real coefficients.

4. Can a cubic equation have more than one positive root?

No, a cubic equation can only have one positive root. This is because a cubic equation can only have a maximum of three roots, and if one is positive, the others must be complex or negative.

5. How can solving a cubic equation with a positive root be useful?

Solving a cubic equation with a positive root can be useful in many real-life applications, such as in engineering, physics, and economics. It can help in finding the maximum or minimum values of a function, optimizing a process, or predicting future trends.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
12
Views
387
  • Precalculus Mathematics Homework Help
Replies
6
Views
546
  • Precalculus Mathematics Homework Help
Replies
13
Views
759
  • Precalculus Mathematics Homework Help
Replies
4
Views
2K
  • Precalculus Mathematics Homework Help
Replies
6
Views
586
  • Precalculus Mathematics Homework Help
Replies
1
Views
876
  • Precalculus Mathematics Homework Help
Replies
22
Views
1K
  • Precalculus Mathematics Homework Help
Replies
20
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
2K
Back
Top