Question

Find the value of $m$ such 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. $-\sqrt{2}$
• C. $-\frac{1}{\sqrt{2}}$
• D. $\sqrt{2}$

Answer 1

Answer: (A), (C)

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

Let $\alpha$ be a common root for the given equations.
Then $\alpha$ will satisfy the equations,
$\Rightarrow 3{\alpha }^2-2m\alpha -4=0\cdots \left(i\right)$
$\Rightarrow {\alpha }^2-4m\alpha +2=0\cdots \left(ii\right)$

Now, solve these equations by cross multiplication method.

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

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

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

$\Rightarrow {\alpha }^2=2\&\alpha =\frac{1}{m}$

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

$\frac{1}{m^2}=2\Rightarrow m^2=\frac{1}{2}\Rightarrow m=\pm \frac{1}{\sqrt{2}}$
$\Rightarrow m=\pm \frac{1}{\sqrt{2}}$