Question

If the roots of the quadratic equation $k^2{x}^2+(kx+1)(x+k)+1=0\forall k\ne 0,-1$ are $\alpha ,\beta$, then the value of ${\alpha }^2{\beta }^2+(\alpha \beta +1)(\alpha +\beta )+1$ will be

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

Given quadratic equation:
$k^2{x}^2+(kx+1)(x+k)+1=0$
$\text{Or}k(k+1)x^2+(k^2+1)x+(k+1)=0$

$\because \alpha ,\beta$ are the roots of the quadratic equation.
$\Rightarrow \alpha +\beta =-\frac{(k^2+1)}{k(k+1)}\cdots \left(i\right)$

$\&\alpha .\beta =\frac{k+1}{k(k+1)}=\frac{1}{k}\cdots \left(ii\right)$

Substituting these values in ${\alpha }^2{\beta }^2+(\alpha \beta +1)(\alpha +\beta )+1$, we get

${\left(\frac{1}{k}\right)}^2+(\frac{1}{k}+1)[-\frac{(k^2+1)}{k(k+1)}]+1$

$={\left(\frac{1}{k}\right)}^2-\frac{(k^2+1)}{k^2}+1$

$=\frac{1-k^2-1+k^2}{k^2}=0$

Hence, ${\alpha }^2{\beta }^2+(\alpha \beta +1)(\alpha +\beta )+1=0$