Problem:
Find the number of real roots of the equation px^2 + qx + r = 0, where p, q, and r are positive real numbers.

Solution:
To find the number of real roots of the quadratic equation px^2 + qx + r = 0, we can consider the discriminant of the equation.

The discriminant, denoted by Δ, is given by the formula: Δ = b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.

In the given equation, the coefficients are p, q, and r, respectively. So we have:
a = p, b = q, c = r

Substituting these values into the formula for the discriminant, we get:
Δ = q^2 - 4pr

Case 1: Δ > 0
If the discriminant is positive, then the quadratic equation has two distinct real roots. However, since p, q, and r are positive real numbers, we have:
q^2 - 4pr > 0

Simplifying the inequality, we get:
q^2 > 4pr

Since all the terms are positive, we can take the square root of both sides without changing the inequality:
q > 2√(pr)

This implies that q is greater than 2 times the square root of the product of p and r. However, since p, q, and r are positive real numbers, this inequality cannot hold true. Therefore, there are no real roots when Δ > 0.

Case 2: Δ = 0
If the discriminant is zero, then the quadratic equation has one real root. However, since p, q, and r are positive real numbers, we have:
q^2 - 4pr = 0

Simplifying the equation, we get:
q^2 = 4pr

Taking the square root of both sides, we get:
q = 2√(pr)

Since q is equal to 2 times the square root of the product of p and r, the equation becomes:
px^2 + 2√(pr)x + r = 0

This is a perfect square trinomial, and it can be factored as:
(p^0.5x + r^0.5)^2 = 0

The only value of x that makes this equation true is:
x = -r^(0.5)/p^(0.5)

Therefore, there is only one real root when Δ = 0.

Case 3: Δ < />
If the discriminant is negative, then the quadratic equation has no real roots. Since p, q, and r are positive real numbers, we have:
q^2 - 4pr < />

This inequality cannot be satisfied, so there are no real roots when Δ < />

Conclusion:
From the three cases discussed above, we can conclude that the given quadratic equation px^2 + qx + r = 0 has no real roots.