Question

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. $3$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is A

$3$

Explanation of correct option.

Step 1: Use the concept of Sum and Product of roots of a Quadratic equation

We know that if $\alpha ,\beta$ are the roots of the quadratic equation $a{x}^2+bx+c=0$ then

$\alpha +\beta =$ $-\frac{b}{a}$ and $\alpha \beta =$$\frac{c}{a}$

Therefore, for the given equation $x^2+px+q=0$,

$\begin{array}{rcl}\tan \left(30°\right)+\tan \left(15°\right)& =& -p....\left(1\right)\\ \tan \left(30°\right)·\tan \left(15°\right)& =& q....\left(2\right)\end{array}$

Step 2: Apply the compound angle formula of tangent

$\tan \left({30}^{\circ }+{15}^{\circ }\right)=\frac{\tan \left(30°\right)+\tan \left(15°\right)}{1-\begin{array}{l}\tan \left(30°\right)·\tan \end{array}\left(15°\right)}...\left(3\right)$

Using $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get

$\tan \left({45}^{\circ }\right)=\frac{-p}{1-q}$

$\begin{array}{rcl}& \Rightarrow & 1=\frac{-p}{1-q}\\ & \Rightarrow & 1-q=-p\\ & \Rightarrow & 1=q-p\\ & \Rightarrow & 1+2=q-p+2\\ & \Rightarrow & 3=2+q-p\\ & \Rightarrow & 2+q-p=3\end{array}$

Hence, the correct answer is an option (A).