Question

Find the value of $m$ so that, the quadratic equations, $3x^2-2mx-4=0$ and $x(x-4m)+2=0$ have a common root.

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

Answer 1

Answer: (A), (B)

$\frac{1}{\sqrt{2}}$

Let $\alpha$ be a common root for the given equations.
Then $\alpha$ will satisfy the equations,

Thus,
$3{\alpha }^2-2m.\alpha -4=0$ ............ (i)
And ${\alpha }^2-4m.\alpha +2=0$ ............ (ii)

Now, solve these equations by cross multiplication method.

$\frac{{\alpha }^2}{-4m-16m}=\frac{\alpha }{-4-6}=\frac{1}{-12m+2m}$

$\frac{{\alpha }^2}{-20m}=\frac{\alpha }{-10}=\frac{1}{-10m}$

$\frac{{\alpha }^2}{-2m}=\frac{\alpha }{-1}=\frac{1}{-m}$

${\alpha }^2=2$ and $\alpha =\frac{1}{m}$

$\Rightarrow {\left(\frac{1}{m}\right)}^2=2$

$\frac{1}{m^2}=2\Rightarrow m^2=\frac{1}{2}\Rightarrow m=\pm \frac{1}{\sqrt{2}}$
Hence, value of $m$ is $\frac{1}{\sqrt{2}}$ and $-\frac{1}{\sqrt{2}}$