Question

lf the equation $({\lambda }^2-5\lambda +6)x^2+({\lambda }^2-3\lambda +2)x+({\lambda }^2-4)=0$ is satisfied by more than two values of $x$, then $\lambda =$

• A. $1$
• B. $-2$
• C. $3$
• D. $2$

Answer 1

The correct option is D $2$

If the quadratic $a{x}^2+bx+c=0$ has more than 2 roots then it is an identify in $x$ and $a=0,b=0,c=0$
$\therefore {\lambda }^2-5\lambda +6=0$

$\lambda =2,\lambda =3$
And ${\lambda }^2-3\lambda +2=0$
$\lambda =2,\lambda =1$
And ${\lambda }^2-4=0$
$\lambda =2,\lambda =-2$
$\therefore$ The common solution is $\lambda =2$