Question

$\text{If}\alpha \beta ,\alpha and\alpha +\beta \text{are the zeroes}\text{of the polynomial,}f\left(x\right)=x^3-3p{x}^2+qx-r,\text{then which of the following relations}\text{is correct.}$

• A.
  $2p^3=pq+r$
• B.
  $p^3=pq+r$
  
• C.
  $2p^3=pq-r$
• D.
  $p^3=pq-r$
  

Answer 1

The correct option is C

$2p^3=pq-r$
$f\left(x\right)=x^3-3p{x}^2+qx-r\text{Its zeroes are}\alpha -\beta ,\alpha ,and\alpha +\beta S=\frac{-b}{a}\Rightarrow \alpha -\beta +\alpha +\alpha +\beta =-\frac{(-3p)}{1}\Rightarrow 3\alpha =3p\Rightarrow \alpha =pf\left(x\right)=x^3-3p{x}^2+qx-r\alpha =p\text{is zero of}f\left(x\right)\Rightarrow f\left(p\right)=0\Rightarrow (p{)}^3-3p(p{)}^2+q\left(p\right)-r=0\Rightarrow p^3-3p^3+pq-r=0\Rightarrow -2p^3+pq-r=0\Rightarrow pq-r=2p^3\therefore 2p^3=pq-r$
So, option c is correct.