If tan `25^(@) and tan 20^(@)` are roots of the quadratic equation `x^(2) + 2px + q = 0`, then 2p - q is equal to

Options

A. `-2`
B. `-1`
C. 0
D. 1

Answer

To solve the problem, we need to find the value of \(2p - q\) given that \(\tan 25^\circ\) and \(\tan 20^\circ\) are the roots of the quadratic equation \(x^2 + 2px + q = 0\).

### Step-by-Step Solution:

1. **Identify the Roots:**
The roots of the quadratic equation are given as:
\[ r_1 = \tan 25^\circ \quad \text{and} \quad r_2 = \tan 20^\circ \]

2. **Use the Sum and Product of Roots:**
For a quadratic equation of the form \(x^2 + bx + c = 0\), the sum and product of the roots can be expressed as:
- Sum of the roots \(r_1 + r_2 = -\frac{b}{a}\)
- Product of the roots \(r_1 \cdot r_2 = \frac{c}{a}\)

Here, \(a = 1\), \(b = 2p\), and \(c = q\).

3. **Calculate the Sum of the Roots:**
\[ r_1 + r_2 = \tan 25^\circ + \tan 20^\circ = -2p \]

4. **Calculate the Product of the Roots:**
\[ r_1 \cdot r_2 = \tan 25^\circ \cdot \tan 20^\circ = q \]

5. **Use the Tangent Addition Formula:**
We know that:
\[ \tan(25^\circ + 20^\circ) = \tan 45^\circ = 1 \]
Using the tangent addition formula:
\[ \tan(25^\circ + 20^\circ) = \frac{\tan 25^\circ + \tan 20^\circ}{1 - \tan 25^\circ \tan 20^\circ} \]
Setting this equal to 1 gives:
\[ 1 = \frac{\tan 25^\circ + \tan 20^\circ}{1 - \tan 25^\circ \tan 20^\circ} \]

6. **Cross-Multiply:**
\[ 1 - \tan 25^\circ \tan 20^\circ = \tan 25^\circ + \tan 20^\circ \]

7. **Substitute the Sum and Product:**
Replacing \(\tan 25^\circ + \tan 20^\circ\) with \(-2p\) and \(\tan 25^\circ \tan 20^\circ\) with \(q\):
\[ 1 - q = -2p \]

8. **Rearranging the Equation:**
\[ 2p = q - 1 \]

9. **Finding \(2p - q\):**
Now, substituting \(2p\) back into the expression:
\[ 2p - q = (q - 1) - q = -1 \]

### Final Answer:
\[ 2p - q = -1 \]