Question

Solve this:
Q.17. If the roots of the quadratic equation $x^2+px+q=0$ are tan 30$°$ and tan 15$°$, respectively then the value of 2 + q - p is
(a) 2
(b) 3
(c) 0
(d) 1

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{equation}\mathrm{is}, \\ {\mathrm{x}}^2+\mathrm{px}+\mathrm{q}=0 \\ \mathrm{Now},\mathrm{sum}\mathrm{of}\mathrm{roots}=-\frac{\mathrm{coefficient}\mathrm{of}\mathrm{x}}{\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2} \\ \Rightarrow \tan 30°+\tan 15°=-\mathrm{p}....\left(1\right) \\ \mathrm{Now},\tan 15°=\tan \left(45°-30°\right) \\ =\frac{\tan 45°-\tan 30°}{1+\tan 45°.\tan 30°} \\ =\frac{1-{\displaystyle \frac{1}{\sqrt{3}}}}{1+{\displaystyle \frac{1}{\sqrt{3}}}} \\ =\frac{\sqrt{3}-1}{\sqrt{3}+1} \\ =\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1} \\ =\frac{3+1-2\sqrt{3}}{3-1} \\ =2-\sqrt{3} \\ \mathrm{Now},\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get} \\ \frac{1}{\sqrt{3}}+2-\sqrt{3}=-\mathrm{p} \\ \Rightarrow \frac{1+2\sqrt{3}-3}{\sqrt{3}}=-\mathrm{p} \\ \Rightarrow \frac{2\sqrt{3}-2}{\sqrt{3}}=-\mathrm{p} \\ \mathrm{Now},\mathrm{product}\mathrm{of}\mathrm{roots}=\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2} \\ \Rightarrow \tan 30°\times \tan 15°=\mathrm{q} \\ \Rightarrow \frac{1}{\sqrt{3}}\times \left(2-\sqrt{3}\right)=\mathrm{q} \\ \Rightarrow \frac{2-\sqrt{3}}{\sqrt{3}}=\mathrm{q} \\ \mathrm{Now},2+\mathrm{q}-\mathrm{p} \\ =2+\frac{2-\sqrt{3}}{\sqrt{3}}+\frac{2\sqrt{3}-2}{\sqrt{3}} \\ =\frac{2\sqrt{3}+2-\sqrt{3}+2\sqrt{3}-2}{\sqrt{3}} \\ =\frac{3\sqrt{3}}{\sqrt{3}} \\ =3 \end{gathered}$$