Probability of a cubic having three real roots.

Question

third degree polynomial is given as follows. f(x)=2x³-ax²+bx+4 a, b coefficients of the equation a,b interval between [-4,4]. what is the probability that all three roots of this equation are real numbers?

Answer 1

If a value is given to a in the stated interval, there is a range of values for b which will allow the equation to have real 3 real roots. That means you could calculate a conditional probability of 3 real roots given a.

I do not believe there is a single answer to this question.For any given value of a there is a range of values for b which will allow the equation to have real roots.

For example if a=4, b must be less than about -1.16, that would make the conditional probability of 3 real roots given a=4 about 2.84/8.

This equation always has one negative real root in the interval given for a and b. If b=a²/6, the equation has a horizontal tangent and as b gets smaller the minimum point goes toward and eventually crosses the x axis.

I cannot help you further and if you get a better answer from another tutor or a better answer in class, I would appreciate knowing about it.

With all due respect to Catherine (above), I do not agree. What she has written concerns the RATIONAL roots of the equation. You have asked for the probability of REAL roots!

Answer 2

1. Use the Rational Root Theorem (p/q) to find all possible real roots:
2. p is the constant term [p = 4], q is the leading coefficient [q = 2]
3. Find all factors of each [p = 4: ±1, ±2, ±4] & [q = 2: ±1, ±2]
4. Then divide each p by q. These are all the possible rational real root [p/q = ±1, ±2, ±4, ±1/2]
5. Probability is the desired outcome divided by the total possible. There are 8 total possible real roots for this polynomial, all 8 lie within the interval [-4, 4]; therefore, P(-4 ≤ x ≤ 4) = 8/8 = 100%