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**Hint:**Every quadratic equation is of the form

\[\begin{align}

& a{{x}^{2}}+bx+c=0 \\

& which\ on\ conversion,\ we\ get \\

& {{x}^{2}}-\left( \dfrac{-b}{a} \right)x+\dfrac{c}{a}=0 \\

\end{align}\]

The coefficient of x gives the sum of the roots of the given equation with an appropriate negative sign that is \[\left( \dfrac{-b}{a} \right)\]. The constant term of the equation will give the product of the two roots of the equation \[\dfrac{c}{a}\] . Now, using these properties, one could get to the solution.

**Complete step by step answer:**

Every quadratic equation is in the form, ${{x}^{2}}-Sx+P=0$(where ‘S’ is the sum of the roots and ‘P’ is the product the roots).

We are given that one of the roots of the given equation \[{{x}^{2}}+\text{ }px\text{ }+\text{ }\left( 1\text{ }\text{- }p \right)\text{ }=\text{ }0\]is (1-p). Let $\alpha $be the other root of the equation.

Sum of the roots$=-p.....................(1)$

Product of the roots=$(1-p)......................(2)$

On comparing the equation given in the question with the general equation, we get,

Sum of the roots ,$S=(1-p)+\alpha ................(3)$

Product of the roots ,$P=(1-p)\alpha ...................(4)$

Equating (1) and (3), we get,

$-p=1-p+\alpha $

$\therefore \alpha =-1$

Hence, one of the roots of the equation \[{{x}^{2}}+\text{ }px\text{ }+\text{ }\left( 1\text{ }\text{- }p \right)\text{ }=\text{ }0\] is -1.

Substituting for $\alpha $ in (4) and equating it with (2), we obtain,

$1-p=-(1-p)$

$2=2p$

$\therefore p=1$

We know that, (1-p) is a root of the equation \[{{x}^{2}}+\text{ }px\text{ }+\text{ }\left( 1\text{ }\text{- }p \right)\text{ }=\text{ }0\],

$\therefore 1-p=1-1=0$

Hence, one of the roots of the equation \[{{x}^{2}}+\text{ }px\text{ }+\text{ }\left( 1\text{ }\text{- }p \right)\text{ }=\text{ }0\]is 0.

$\therefore $ The roots of the equation \[{{x}^{2}}+\text{ }px\text{ }+\text{ }\left( 1\text{ }\text{- }p \right)\text{ }=\text{ }0\]are 0 and -1.

**So, the correct answer is “Option a”.**

**Note:**Alternate method:

We have,

\[{{x}^{2}}+\text{ }px\text{ }+\text{ }\left( 1\text{ }\text{- }p \right)\text{ }=\text{ }0......................(1)\]

We are given (1 – p) is a root of quadratic equation\[{{x}^{2}}+\text{ }px\text{ }+\text{ }\left( 1\text{ }\text{- }p \right)\text{ }=\text{ }0\] .

Hence, on substituting $x$ by (1 – p), we get,

\[{{(1-p)}^{2}}+\text{ }p(1-p)\text{ }+\text{ }\left( 1\text{ }\text{ }p \right)\text{ }=\text{ }0\]

Taking (1-p) common for all the terms, we obtain,

\[(1-p)\left( (1-p)+p+1 \right)=0\]

\[(1-p)\left( 2 \right)=0\]

Since, 2 is a constant and cannot take the value of zero,

\[1-p=0\]

\[\therefore p=1\]

Substituting p=1 in equation (1), we obtain,

\[{{x}^{2}}+\text{ (1)}x\text{ }+\text{ }\left( 1\text{ }\text{ 1} \right)=0\]

\[{{x}^{2}}+\text{ }x\text{ }=0\]

$x(x+1)=0$

Hence,

$x=0.............(2)$

$x+1=0$

$x=-1...............(3)$

$\therefore $ From (2) and (3), the zeroes of the equation \[{{x}^{2}}+\text{ }px\text{ }+\text{ }\left( 1\text{ }\text{- }p \right)\text{ }=\text{ }0\]are 0 and -1.

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