Question

The equation $x^2-6x+8+\lambda (x^2-4x+3)=0,\forall \lambda \in R-\{-1\}$ has

• A. real and unequal roots for all $\lambda \in R-\{-1\}$
• B. real and equal roots for $\lambda <0$
• C. real and equal roots for $\lambda >0$
• D. real and equal roots for $\lambda =0$

Answer 1

The correct option is A real and unequal roots for all $\lambda \in R-\{-1\}$

$x^2-6x+8+\lambda (x^2-4x+3)=0$
$\Rightarrow (1+\lambda )x^2-2(3+2\lambda )x+(8+3\lambda )=0$

Discriminant of above quadratic equation is,
$\Delta =4(3+2\lambda {)}^2-4(1+\lambda \left)\right(8+3\lambda )$
$=4\left[\right(3+2\lambda {)}^2-(1+\lambda )(8+3\lambda )]$
$=4[4\lambda +12\lambda +9-(3{\lambda }^2+11\lambda +8\left)\right]$
$=4({\lambda }^2+\lambda +1)$

Now, since discriminant of quadratic ${\lambda }^2+\lambda +1=0$ is negative,
$\Delta >0$ for all $\lambda \in R-\{-1\}$
Hence, roots are real and unequal.
Hence, first option is correct.