Question

The value of $m$ for which the equations $3x^2-2mx-4=0$ and $x^2-4mx+2=0$ have a common root is

• A. $\frac{1}{\sqrt{2}}$
• B. $-\frac{1}{\sqrt{2}}$
• C. $\sqrt{2}$
• D. $-\sqrt{2}$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{1}{\sqrt{2}}$
B $-\frac{1}{\sqrt{2}}$

Given,
$3x^2-2mx-4=0$ and $x^2-4mx+2=0$ have a common root.
Let $\alpha$ be the common root.
$3{\alpha }^2-2m\alpha -4=0$
${\alpha }^2-4m\alpha +2=0$
By Cramer's rule,
$\frac{{\alpha }^2}{-4m-16m}=\frac{\alpha }{-4-6}=\frac{1}{-12m+2m}$
$\frac{{\alpha }^2}{-20m}=\frac{\alpha }{-10}=\frac{1}{-10m}$
$\alpha =2m,\alpha =\frac{1}{m}$
$\Rightarrow 2m=-\frac{1}{m}$
$\Rightarrow m^2=\frac{1}{2}$
$\Rightarrow m=\pm \frac{1}{\sqrt{2}}$
Hence, options 'A' and 'B' are correct.