Need proof that a cubic equation has at least one real root

In summary, it is possible to prove that a cubic equation has at least one real root by using the continuity and asymptotic limits of real polynomials, as well as the Rolle theorem. Alternatively, one can also show that complex roots occur in conjugate pairs, leaving an odd number of unpaired real roots.
  • #1
lmamaths
6
0
Hi,

A cubic equation has at least one real root.
Can someone help me to prove this?

Thx!
LMA
 
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  • #2
U need calculus.It's not difficult.Use the continuitiy on [itex] \mathbb{R} [/itex] and evaluate the 2 asymptotic limits.Then use the Rolle (IIRC) theorem...

Daniel.
 
  • #3
I don't think you necessarily need that.

If w is a complex root of f(x) a real polynomial (ie one with real coeffs), then so is w* the conjugate of w, this means that the real poly (x-w)(x-w*) divides f(x) over the reals. Hence complex roots occur in conjugate pairs. A cubic has 3 (possibly complex) roots, so pairing up the complex ones (or any poly of odd degree) means there must be an odd number of unpaired real roots left.
 

Related to Need proof that a cubic equation has at least one real root

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.

2. How do you know that a cubic equation has at least one real root?

This can be determined by the fundamental theorem of algebra, which states that a polynomial equation of degree n has exactly n complex roots. Since a cubic equation has a degree of 3, it must have at least one real root.

3. Can you provide an example of a cubic equation with at least one real root?

Sure, one example is x^3 - 2x + 5 = 0. This equation has a real root of approximately 1.38.

4. How can you prove that a cubic equation has at least one real root?

One way to prove this is by using the intermediate value theorem, which states that if a continuous function has different signs at two points, there must be at least one root between those two points. The cubic equation can be represented as a continuous function, and by finding two points with different signs, we can prove that there is at least one real root.

5. Are there any special cases where a cubic equation may not have a real root?

Yes, if all three roots of the cubic equation are complex numbers, then there will be no real roots. This can happen if the discriminant, b^2 - 4ac, is negative.

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