Let us consider the quadratic equation: ax˛ + bx + c = 0.

We assign values to the coefficients a, b and c by throwing a die.

What is the probability that the equation will have real roots ?

(In reply to Quadratic Equation Coefficients by Gordon Steel)

What you have said is correct, the roots of a quadratic equation can be real or imaginary, but it is not true that integral coefficients ensure that the roots will be integral, rational, or even real at all.

If you have a quadratic equation ax²+bx+c=0, then the quadratic states that the values of x are shown by:

     -b ± √(b²-4ac)


x =  --------------


           2a

This kind of equation can have one or two real roots, or two imaginary roots.
The part under the radical sign, b²-4ac, called the discriminant, determines the types of roots the equation will have.

If the discriminant is zero, then the square root drops out and you have one rational root.

If the discriminant is a perfect square, you will have two rational roots.

A rational root will be integral iff the denominator, 2a, is a 'clean' factor of the numerator.

If the discriminant is positive, but not a perfect square, the equation has two irrational roots.

If the discriminant is negative, then you have the square root of a negative number, which is a complex (imaginary) number, involving (as you said) i, which is defined as a number that, when squared, yields -1. For example, √(-4) could be rewritten as √(4×-1)=√2×√(-1)=2i.

To solve this problem, you need to find the probability that for the coefficients determined by the dice (a, b, and c), the value of the discriminant will be nonnegative. If 4ac is greater than b², which is certainly possible with values ranging from 1 to 6 for all three coefficients, then the discriminant will be negative and the roots will be imaginary.

Posted by DJ on 2003-08-20 23:33:13