Question

If (1 – p) is a root of quadratic equation \[{{x}^{2}}+\text{ }px\text{ }+\text{ }\left( 1\text{ }\text{- }p \right)\text{ }=\text{ }0\] , then both the roots are:-
a)0, -1
b)-1, 1
c)0, 1
d)-1, 2

Answer 1

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.