If $$(1 - p)$$ is a root of quadratic equation $$x^2 + px + (1 - p) = 0$$, then its roots are

Let the given quadratic equation be:

$$ x^2 + px + (1 - p) = 0 $$

Since $$ 1 - p $$ is a root of this equation, it must satisfy the equation. Substitute $$ x = 1 - p $$ into the equation:

$$ (1 - p)^2 + p(1 - p) + (1 - p) = 0 $$

Factor out the common term $$ 1 - p $$ from the expression:

$$ (1 - p) ((1 - p) + p + 1) = 0 $$

Simplify the terms inside the second parentheses:

$$ (1 - p) (1 - p + p + 1) = 0 $$

$$ (1 - p) (2) = 0 $$

Divide both sides by 2 to find the value of $$ p $$:

$$ 1 - p = 0 $$

$$ p = 1 $$

Substitute the value of $$ p = 1 $$ back into the original quadratic equation:

$$ x^2 + (1)x + (1 - 1) = 0 $$

$$ x^2 + x = 0 $$

Factor the simplified quadratic equation to find its roots:

$$ x(x + 1) = 0 $$

This gives two possible values for $$ x $$:

$$ x = 0 $$

$$ x + 1 = 0 $$

$$ x = -1 $$

Final Answer:

The roots of the quadratic equation are 0 and -1.