Question

If the equations $4x^2-x-1=0$ and $3x^2+(\lambda +\mu )x+\lambda -\mu =0$ have a root common, then the rational values of $\lambda$ and $\mu$ are

• A. $\lambda =-\frac{3}{4}$
• B. $\lambda =11$
• C. $\mu =\frac{1}{4}$
• D. $\mu =0$

Answer 1

Answer: (A), (D)

$\lambda =-\frac{3}{4}$
$\mu =0$

Given,
$4x^2-x-1=0$ and $3x^2+(\lambda +\mu )x+\lambda -\mu =0$
$4x^2-x-1=0=0$
$\Rightarrow$$x=$$\frac{-(-1)\pm \sqrt{1-4(-4)}}{2\times 4}$
$=$$\frac{1\pm \sqrt{17}}{8}$
Here, the roots are irrational.
If there is exactly one common root $\lambda$ and $\mu$ cannot have rational values as $-\left(\frac{\lambda +\mu }{3}\right)$ and $\frac{\lambda -\mu }{3}$ which represent the sum of the roots and product of the roots will become irrational if the irrational roots are not conjugate.
So, for $\lambda ,\mu$ to have rational values both the roots should be common i.e.,they should have identical equations.
$\Rightarrow$$\frac{4}{3}=\frac{-1}{\lambda +\mu }=\frac{-1}{\lambda -\mu }$
$\lambda +\mu =-\frac{3}{4}$
$\lambda -\mu =-\frac{3}{4}$
From both the equations,
$\lambda =-\frac{3}{4}$ and $\mu =0$
Hence, options 'A' and 'D' are correct.