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# Roots of equation $x^2+px-q=0$ are $\tan 30^{\large\circ}$ and $\tan 15^{\large\circ}$,then value of $2-q-p$ is :

$\begin{array}{1 1}(A)\;0&(B)\;1\\(C)\;2&(D)\;3\end{array}$

Can you answer this question?

$-p=\tan 30^{\large\circ}+\tan 15^{\large\circ}$
$-q=\tan 30^{\large\circ}.\tan 15^{\large\circ}$
$\tan 45^{\large\circ}=\tan(30+15)^{\large\circ}=\large\frac{\tan 30^{\large\circ}+\tan 15^{\large\circ}}{1-\tan 30^{\large\circ}\tan 15^{\large\circ}}$
$\Rightarrow 1=-\large\frac{p}{1+q}$$\Rightarrow -p-q=1$
$\Rightarrow 2-p-q=2+1=3$
Hence (D) is the correct answer.
answered Apr 15, 2014