Question

Number of real values of $\lambda$ so the equation $x^2-3x+2\lambda =0$ and $x^2-4\lambda x+3=0$ has exactly one root common

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is A $1$

Let $\alpha$ be the common root

Then ${\alpha }^2-3\alpha +2\lambda =0$

and ${\alpha }^2-4\alpha \lambda +3=0$

Therefore, $\frac{{\alpha }^2}{-9+8{\lambda }^2}=\frac{\alpha }{2\lambda -3}=\frac{1}{-4\lambda +3}$

$\Rightarrow {\alpha }^2=\frac{8{\lambda }^2-9}{2\lambda -3}$ and $\alpha =\frac{2\lambda -3}{-4\lambda +3}$

$\Rightarrow {\left(\frac{2\lambda -3}{-4\lambda +3}\right)}^2=\frac{8{\lambda }^2-9}{2\lambda -3}$

$\Rightarrow \frac{4{\lambda }^2+9-12\lambda }{-4\lambda +3}=8{\lambda }^2-9$

$\Rightarrow 4{\lambda }^2-12\lambda +9=-32{\lambda }^3+36\lambda -27$

$\Rightarrow 32{\lambda }^3-20{\lambda }^2-48\lambda +36=0$

Let $p\left(\lambda \right)=32{\lambda }^3-20{\lambda }^2-48\lambda +36$

Since $p\left(1\right)=0$ therefore by factor theorem $\lambda -1$ is a factor of $p\left(\lambda \right)$

On dividing $p\left(\lambda \right)$ by $\lambda -1$ we get

$p\left(\lambda \right)=\left(\lambda -1\right)\left(32{\lambda }^2+12\lambda +36\right)$

$=4\left(\lambda -1\right)\left[8{\lambda }^2+3\lambda +9\right]$

Since Discriminant of ${\lambda }^2+3\lambda +9$ i.e $D=$$9-4\times 1\times 9<0$

Therefore it has two imaginary roots

Hence $p\left(\lambda \right)$ has one real root and two imaginary roots