Question

If the roots of the quadratic equation $mx^2 - nx + k = 0$ are tan $33^{\circ}$ and $\tan\, 12^{\circ}$ then the value of $\frac{2m+n+k}{m}$ is equal to

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The Correct Option is D

Given, quadratic equation is
$m x^{2}-n x+k=0$
Roots of the equation are $\tan 33^{\circ}$ and $\tan 12^{\circ} .$
$\therefore \tan 33^{\circ}+\tan 12^{\circ}=\frac{n}{m} \ldots (i)$
and $\tan 33^{\circ} \times \tan 12^{\circ}=\frac{k}{m} \ldots(ii)$
Value of $\frac{2 m+n+k}{m}$ is
$\frac{2 m+n+k}{m}= \frac{2 m}{m}+\frac{n}{m}+\frac{k}{m} $
$=2+\left(\tan 33^{\circ}+\tan 12^{\circ}\right) $
$+\left(\tan 33^{\circ} \times \tan 12^{\circ}\right) \ldots (iii)$
Let $\left(\tan 45^{\circ}\right)=\tan \left(33^{\circ}+12^{\circ}\right)$
$\Rightarrow 1=\frac{\tan 33^{\circ}+\tan 12^{\circ}}{1-\tan 33^{\circ} \tan 12^{\circ}}$
$\Rightarrow 1-\tan 33^{\circ} \tan 12^{\circ}=\tan 33^{\circ}+\tan 12^{\circ}$
$\Rightarrow \tan 33^{\circ}+\tan 12^{\circ}$
$+\tan 33^{\circ} \times \tan 12^{\circ}=1 \ldots (iv)$
By putting the value from E (iv) into E (iii)
$\frac{2 m+n+k}{m}=2+1=3$