Introduction:
In this question, we are asked to find the condition for which the roots of the given equation are real.

Given Equation:
x² (2P-1)x P²= 0

Determining Real Roots:
For the given equation to have real roots, the discriminant of the equation should be greater than or equal to zero.

Deriving Discriminant:
The discriminant of the given equation can be derived as follows:

D = b² - 4ac
Where a, b, and c are the coefficients of x², x, and constant terms respectively.

In the given equation, a = P², b = (2P - 1), and c = 0

Substituting:
D = (2P - 1)² - 4P² * 0

Simplifying:
D = 4P² - 4P + 1 - 0

Simplifying Further:
D = 4P² - 4P + 1

Condition for Real Roots:
For the given equation to have real roots, the discriminant (D) should be greater than or equal to zero.

Solving the Inequality:
4P² - 4P + 1 ≥ 0

Factorizing:
(2P - 1)² ≥ 0

Conclusion:
The above inequality is always true for any value of P. Therefore, the roots of the given equation x² (2P-1)x P²= are always real.

X^2+(2p+1)x+p^2=0
a=I,b=2p-1,c=p^2
D=b^2-4ac
D=(2p-1)^2-4(1)(p^2)
D=4p^2-4p+1-4p^2
D=-4p+1
Equating D to 0
-4p+1=0
-4p=-1
4p=1
p=1/4
The roots of the equation are real if p>1/4
Option C