Prove that if p and q are rational and p is not equal to 0, the roots of the quadratic equations p x(square) + 2qx -p +2q = 0.
To prove that if p and q are rational numbers and p is not equal to 0, the quadratic equation p(x^2) + 2qx - p + 2q = 0 has rational roots, we need to show that the discriminant of the quadratic equation is a perfect square.
The discriminant of a quadratic equation ax^2 + bx + c = 0 is given by b^2 - 4ac.
In our case, the quadratic equation is p(x^2) + 2qx - p + 2q = 0, so a = p, b = 2q, and c = -p + 2q.
Calculating the discriminant:
Discriminant (D) = (2q)^2 - 4(p)(-p + 2q)
= 4q^2 + 4p^2 - 8pq
= 4(q^2 + p^2 - 2pq)
Now, we want to show that the discriminant D is a perfect square. To do this, we need to express D as the square of a rational number, say r, where r = m/n, and m and n are integers with no common factors other than 1.
So, D = r^2
= (m/n)^2
= m^2 / n^2
Now, let's express q, p, and r in terms of integers:
q = a/b (where a and b are integers with no common factors),
p = c/d (where c and d are integers with no common factors),
r = m/n (where m and n are integers with no common factors).
Substituting these expressions into the expression for D, we have:
4(q^2 + p^2 - 2pq) = (m^2 / n^2)
Rearranging the equation:
4(q^2 + p^2 - 2pq) = m^2 / n^2
Multiplying both sides by n^2:
4n^2(q^2 + p^2 - 2pq) = m^2
Since n^2 is an integer, the left-hand side must also be an integer. Therefore, m^2 has to be divisible by 4.
For m^2 to be divisible by 4, m must be divisible by 2. Let m = 2k, where k is an integer.
Substituting this back into the equation, we get:
4n^2(q^2 + p^2 - 2pq) = (2k)^2
n^2(q^2 + p^2 - 2pq) = k^2
Now, since k^2 is a perfect square, and the left-hand side has no factor other than n (which is rational), it follows that q^2 + p^2 - 2pq must also be a perfect square.
Therefore, if p and q are rational numbers, and p is not equal to 0, the roots of the quadratic equation p(x^2) + 2qx - p + 2q = 0 will be rational.