Question

If the roots of the quadratic equation ${{x}^{2}}+px+q=0$ are $\tan 30{}^\circ $ and $\tan 15{}^\circ $ respectively, then the value of $2+q-p$ is
1) $3$
2) $0$
3) $1$
4) $2$

Answer 1

Hint: In this problem we need to calculate the value of $2+q-p$ where $\tan 30{}^\circ $ and $\tan 15{}^\circ $ are roots of the quadratic equation ${{x}^{2}}+px+q=0$. Here we will use the relation between the coefficients of the quadratic equation and roots of the quadratic equation. For this we will compare the given quadratic equation with $a{{x}^{2}}+bx+c=0$. Here we will have the value of $p$ and $q$ , after having the values calculate the value of $q-p$ . Use the appropriate trigonometric formulas and values to simplify the value of $q-p$. After having the value of $q-p$, simply calculate the required value.

Complete step-by-step solution:
The quadratic equation is ${{x}^{2}}+px+q=0$.
Compare the above quadratic equation with standard quadratic equation $a{{x}^{2}}+bx+c=0$.
$a=1$ , $b=p$ and $c=q$ .
If $\alpha $ , $\beta $ are the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$, then the relation between the roots and coefficients of the quadratic equation are given by
$\alpha +\beta =-\dfrac{b}{a}$ and $\alpha \beta =\dfrac{c}{a}$ .
In the problem they have mentioned that $\tan 30{}^\circ $ and $\tan 15{}^\circ $ are roots of the quadratic equation ${{x}^{2}}+px+q=0$. So we can write
$\begin{align}
  & \tan 30{}^\circ +\tan 15{}^\circ =-\dfrac{p}{1} \\
 & \Rightarrow \tan 30{}^\circ +\tan 15{}^\circ =-p \\
\end{align}$ and $\begin{align}
  & \tan 30{}^\circ .\tan 15{}^\circ =\dfrac{q}{1} \\
 & \Rightarrow \tan 30{}^\circ .\tan 15{}^\circ =q \\
\end{align}$
Now the value of $q-p$ will be
$q-p=\tan 30{}^\circ .\tan 15{}^\circ +\tan 30{}^\circ +\tan 15{}^\circ ...\left( \text{i} \right)$
From the trigonometric formula $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A.\tan B}$ , we can write the value of $\tan 30{}^\circ +\tan 15{}^\circ $ as $\tan 30{}^\circ +\tan 15{}^\circ =\left[ \tan \left( 30{}^\circ +15{}^\circ \right) \right]\left[ 1-\tan 30{}^\circ .\tan 15{}^\circ \right]$ . Substituting this value in the equation $\left( \text{i} \right)$ , then we will have
$\begin{align}
  & q-p=\tan 30{}^\circ .\tan 15{}^\circ +\left[ \tan \left( 30{}^\circ +15{}^\circ \right) \right]\left[ 1-\tan 30{}^\circ .\tan 15{}^\circ \right] \\
 & \Rightarrow q-p=\tan 30{}^\circ .\tan 15{}^\circ +\left[ \tan 45{}^\circ \left( 1-\tan 30{}^\circ .\tan 15{}^\circ \right) \right] \\
\end{align}$
We know that the value of $\tan 45{}^\circ =1$ . substituting this value in the above equation, then we will get
$\begin{align}
  & q-p=\tan 30{}^\circ .\tan 15{}^\circ +\left[ 1\left( 1-\tan 30{}^\circ .\tan 15{}^\circ \right) \right] \\
 & \Rightarrow q-p=\tan 30{}^\circ .\tan 15{}^\circ +1-\tan 30{}^\circ .\tan 15{}^\circ \\
 & \Rightarrow q-p=1 \\
\end{align}$
Add $2$ on both sides of the above equation, then we will have
$\begin{align}
  & 2+q-p=1+2 \\
 & \therefore 2+q-p=3 \\
\end{align}$
Hence option 1 is the correct answer.

Note: We can also solve the problem in another method. In this method we will first calculate the values of $p$ and $q$ by using the trigonometric formula $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A.\tan B}$ individually. After simplifying the values of $p$ and $q$ we can calculate the required value. But it is some lengthy process and consumes extra time.