If `sin theta, cos theta` are the roots of `6x^(2)-px+1=0`, then `p^(2)` =

Options

A. 40
B. 48
C. 36
D. 24

Answer

To solve the problem where `sin theta` and `cos theta` are the roots of the quadratic equation `6x^2 - px + 1 = 0`, we can follow these steps:

### Step 1: Identify the relationships from the roots
Given that `sin theta` and `cos theta` are the roots of the quadratic equation, we can use Vieta's formulas:
- The sum of the roots (sin theta + cos theta) is given by:
\[ \text{Sum of roots} = \frac{p}{6} \]
- The product of the roots (sin theta * cos theta) is given by:
\[ \text{Product of roots} = \frac{1}{6} \]

### Step 2: Set up equations
From Vieta's formulas, we have:
1. \( \sin \theta + \cos \theta = \frac{p}{6} \) (Equation 1)
2. \( \sin \theta \cdot \cos \theta = \frac{1}{6} \) (Equation 2)

### Step 3: Use the Pythagorean identity
We know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \quad \text{(Equation 3)} \]

### Step 4: Express sin²θ + cos²θ in terms of the sum and product of the roots
Using the identity:
\[ \sin^2 \theta + \cos^2 \theta = (\sin \theta + \cos \theta)^2 - 2 \sin \theta \cos \theta \]
Substituting from Equations 1 and 2:
\[ 1 = \left(\frac{p}{6}\right)^2 - 2 \cdot \frac{1}{6} \]

### Step 5: Simplify the equation
Now, we will simplify the equation:
\[ 1 = \frac{p^2}{36} - \frac{2}{6} \]
\[ 1 = \frac{p^2}{36} - \frac{1}{3} \]
To eliminate the fraction, multiply the entire equation by 36:
\[ 36 = p^2 - 12 \]
Adding 12 to both sides gives:
\[ p^2 = 48 \]

### Conclusion
Thus, the value of \( p^2 \) is:
\[ \boxed{48} \]