Suppose p and q are odd integers.

(a) Show that the quadratic equation x2 + px + q = 0 has no rational roots.
(A number á is a “root” of that equation if: á2 + pá + q = 0. A number is rational if it is expressible as
m/n for some integers m and n. )
(b) Does this result generalize to equations of the type xn + px + q = 0 ?

(a) To show that the quadratic equation x^2 + px + q = 0 has no rational roots, we can use the Rational Root Theorem. According to the theorem, if a rational number r is a root of the equation, then r must be of the form r = m/n, where m is a factor of the constant term q and n is a factor of the leading coefficient 1.

In this case, we have q as an odd integer. The only possible factors of an odd integer are odd integers themselves. Similarly, we have 1 as the leading coefficient, which has 1 as its only factor.

So, any rational root of the form m/n would require m and n to be odd integers. Now let's consider the equation when we substitute r = m/n into it:

(m/n)^2 + p(m/n) + q = 0
(m^2 + pmn + qn^2) / n^2 = 0

Since m, n, and q are odd integers, m^2, p, mn, and qn^2 are also odd. Therefore, the numerator (m^2 + pmn + qn^2) would be an odd number.

However, for this equation to have a rational root, the numerator should be divisible by n^2. Since odd numbers cannot be divided by an even number (n^2 being even), we reach a contradiction.

Thus, x^2 + px + q = 0 has no rational roots when p and q are odd integers.

(b) No, this result does not generalize to equations of the type x^n + px + q = 0, where n is any positive integer greater than or equal to 3.

The generalization of this problem is known as the Rational Root Theorem for polynomials. For a polynomial equation of the form x^n + px + q = 0, where n is a positive integer greater than or equal to 3, the Rational Root Theorem states that if a rational number r is a root of the equation, then r must be of the form r = m/n, where:
- m is a factor of the constant term q,
- n is a factor of the leading coefficient 1, and
- m and n are coprime (i.e., their greatest common divisor is 1).

Thus, for the equation x^n + px + q = 0 to have rational roots, the constant term q must have factors that are odd integers, and the leading coefficient must be 1.

However, in the case of odd integers p and q, a quadratic equation is a special case of this general equation where n = 2. The Rational Root Theorem states that for n = 2, the only potential rational roots are of the form m/n, where m is a factor of the constant term q and n is a factor of the leading coefficient 1.

Since we already showed in part (a) that x^2 + px + q = 0 has no rational roots when p and q are odd integers, we can conclude that for higher powers of x (n > 2), the equation x^n + px + q = 0 will also not have any rational roots.

To prove that the quadratic equation x^2 + px + q = 0 has no rational roots, we can use the Rational Root Theorem. The Rational Root Theorem states that if a polynomial equation such as this one has a rational root, it must be of the form p/q, where p is a factor of the constant term (in this case, q) and q is a factor of the leading coefficient (in this case, 1).

(a)
1. Assume for contradiction that the quadratic equation x^2 + px + q = 0 has a rational root, let's say r, where r is in the form p/q.
2. Substitute r into the equation: (p/q)^2 + p(p/q) + q = 0.
3. Simplify to get p^2/q^2 + p^2/q + q = 0.
4. Multiply the equation by q^2 to eliminate the denominators: p^2 + p^2(q/q)q + q^3 = 0.
5. Simplify further: p^2 + p^3 + q^3 = 0.

Now, let's consider the properties of odd integers p and q:
- An odd number squared is always odd (p^2 is odd).
- An odd number multiplied by an odd number is always odd (p^2 * p = p^3 is odd).
- The sum of two odd numbers is always even (p^2 + p^3 is even).
- The cube of an odd number is always odd (q^3 is odd).

Therefore, the equation p^2 + p^3 + q^3 = 0 cannot hold true for odd integers p and q because an even number (p^2 + p^3) cannot equal an odd number (q^3). Thus, our assumption that the quadratic equation has a rational root is false.

(b) The result does not necessarily generalize to equations of the type x^n + px + q = 0, where n is an arbitrary positive integer.

For higher degree equations, the Rational Root Theorem does not guarantee that there are no rational roots. It only provides a list of possible rational roots to check. If there are no rational roots among those possibilities, it does not necessarily mean there are no rational roots at all.

To determine if a higher degree equation has rational roots, one can use various techniques such as factoring, synthetic division, or applying the Rational Root Theorem to test possible rational roots. However, in general, it is not as straightforward as the case of a quadratic equation.

So, to summarize, for the given quadratic equation x^2 + px + q = 0, it has no rational roots when p and q are odd integers. However, this result does not hold true in general for equations of the type x^n + px + q = 0.