Question

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

• A. $0$
• B. $1$
• C. $3$
• D. $2$

Answer 1

The correct option is C $3$

Given, quadratic equation is
$m{x}^2-nx+k=0$

Roots of the equation are $\tan {33}^o$ and $\tan {12}^o$

$\therefore \tan {33}^o+\tan {12}^o=\frac{n}{m}...\left(i\right)$

$\tan {33}^o\times \tan {12}^o=\frac{k}{m}...\left(ii\right)$

value of $\frac{2m+n+k}{m}$ is

$\frac{2m+n+k}{m}=\frac{2m}{m}+\frac{n}{m}+\frac{k}{m}$

$=2+(\tan {33}^o+\tan {12}^o)+(\tan {33}^o\times \tan {12}^o)...\left(iii\right)$

Let $\left(\tan {45}^o\right)=\tan ({33}^o+{12}^o)$

$\Rightarrow 1=\frac{\tan {33}^o+\tan {12}^o}{1-\tan {33}^o\tan {12}^o}$

$\Rightarrow 1-\tan {33}^o\tan {12}^o=\tan {33}^o+\tan {12}^o$

$\Rightarrow \tan {33}^o+\tan {12}^o+\tan {33}^o\times \tan {12}^o=\mathrm{1...}\left(iv\right)$

By putting the value from Eq(iv) into Eq.(iii)
$\frac{2m+n+k}{m}==2+1=3$