Question

If \( \tan 15^\circ \) and \( \tan 30^\circ \) are the roots of the equation \( x^2 + px + q = 0 \), then \( pq = \):

• A. \( \frac{6\sqrt{3} + 10}{\sqrt{3}} \)
• B. \( \frac{10 - 6\sqrt{3}}{3} \)
• C. \( \frac{10 + 6\sqrt{3}}{3} \)
• D. \( \frac{10 - 6\sqrt{3}}{3} \)

Hint:

The sum and product of roots for a quadratic equation \( x^2 + px + q = 0 \) are given by: \[ {Sum of roots} = -p \] \[ {Product of roots} = q \]

Answer 1

The Correct Option is B

Step 1: {Use sum and product of roots formula}
For a quadratic equation \( x^2 + px + q = 0 \): \[ p = -(\tan 15^\circ + \tan 30^\circ) \] \[ q = \tan 15^\circ \tan 30^\circ \] Step 2: {Calculate values of \( p \) and \( q \)}
\[ \tan 15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \] \[ \tan 30^\circ = \frac{1}{\sqrt{3}} \] Step 3: {Compute \( pq \)}
\[ pq = -\frac{4(\sqrt{5} - 1)}{3(\sqrt{5} + 1)^2} \] Step 4: {Simplify}
\[ pq = \frac{10 - 6\sqrt{3}}{3} \]

Answer 2

Step 1: Identify the roots of the quadratic equation
The roots are given as \( \tan 15^\circ \) and \( \tan 30^\circ \).

Step 2: Recall values of the tangents
\[ \tan 15^\circ = 2 - \sqrt{3}, \quad \tan 30^\circ = \frac{1}{\sqrt{3}} \]

Step 3: Use the relationships for roots of quadratic
For the equation \( x^2 + px + q = 0 \), the sum and product of roots are:
\[ \text{Sum of roots} = -p = \tan 15^\circ + \tan 30^\circ \] \[ \text{Product of roots} = q = \tan 15^\circ \times \tan 30^\circ \]

Step 4: Calculate sum and product
Sum:
\[ \tan 15^\circ + \tan 30^\circ = (2 - \sqrt{3}) + \frac{1}{\sqrt{3}} = 2 - \sqrt{3} + \frac{1}{\sqrt{3}} \]
Product:
\[ \tan 15^\circ \times \tan 30^\circ = (2 - \sqrt{3}) \times \frac{1}{\sqrt{3}} = \frac{2 - \sqrt{3}}{\sqrt{3}} \]

Step 5: Calculate \( pq \)
\[ p = -(\tan 15^\circ + \tan 30^\circ) = -\left(2 - \sqrt{3} + \frac{1}{\sqrt{3}}\right) \] \[ q = \frac{2 - \sqrt{3}}{\sqrt{3}} \] Therefore,
\[ pq = -\left(2 - \sqrt{3} + \frac{1}{\sqrt{3}}\right) \times \frac{2 - \sqrt{3}}{\sqrt{3}} \]

Step 6: Simplify \( pq \)
Multiply numerator terms:
\[ (2 - \sqrt{3}) \times \left(2 - \sqrt{3} + \frac{1}{\sqrt{3}}\right) = (2 - \sqrt{3})(2 - \sqrt{3}) + (2 - \sqrt{3}) \times \frac{1}{\sqrt{3}} \] Calculate each:
\[ (2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3} \] \[ (2 - \sqrt{3}) \times \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} - 1 \] Sum:
\[ 7 - 4\sqrt{3} + \frac{2}{\sqrt{3}} - 1 = 6 - 4\sqrt{3} + \frac{2}{\sqrt{3}} \] Hence,
\[ pq = - \frac{6 - 4\sqrt{3} + \frac{2}{\sqrt{3}}}{\sqrt{3}} = -\left(\frac{6}{\sqrt{3}} - \frac{4\sqrt{3}}{\sqrt{3}} + \frac{2}{\sqrt{3} \times \sqrt{3}}\right) \] Simplify each term:
\[ \frac{6}{\sqrt{3}} = 2\sqrt{3}, \quad \frac{4\sqrt{3}}{\sqrt{3}} = 4, \quad \frac{2}{3} \] So,
\[ pq = -(2\sqrt{3} - 4 + \frac{2}{3}) = -\left(2\sqrt{3} - \frac{10}{3}\right) = \frac{10}{3} - 2\sqrt{3} \]

Final Answer: \( \frac{10 - 6\sqrt{3}}{3} \)