Question

If the roots of the quadratic equation $m{x}^2-nx+k=0$ 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
• E. 4

Answer 1

Answer: (D)

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