Question

The value of $m$ so that the equation $3x^2-2mx-4=0$ and $x(x-4m)+2=0$ may have a common root is

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

Answer 1

Answer: (A), (B)

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

Let $\alpha$ be the common root
Then $3{\alpha }^2-2m\alpha -4=0$ and ${\alpha }^2-4m\alpha +2=0$
By cross-multiplication, we get
$\frac{{\alpha }^2}{-4m-16m}=\frac{\alpha }{-4-6}=\frac{1}{-12m+2m}$

$\Rightarrow \frac{{\alpha }^2}{-20m}=\frac{\alpha }{-10}=\frac{1}{-10m}\Rightarrow \frac{{\alpha }^2}{2m}=\frac{\alpha }{1}=\frac{1}{m}$
From first two terms, $\alpha =2m$ and from the last two terms, $\alpha =\frac{1}{m}$

$\therefore 2m=\frac{1}{m}$

$\therefore 2m^2=1$

$\therefore m=\pm \frac{1}{\sqrt{2}}$