Quadratic equation `8 "sec"^(2) theta - 6 "sec" theta + 1 = 0` has

Options

A. exactly 2 roots
B. exactly 4 roots
C. infinitely many roots
D. no roots

Answer

To solve the quadratic equation \( 8 \sec^2 \theta - 6 \sec \theta + 1 = 0 \), we can follow these steps:

### Step 1: Substitute \( \sec \theta \) with \( y \)
Let \( y = \sec \theta \). Then, the equation becomes:
\[ 8y^2 - 6y + 1 = 0 \]

### Step 2: Use the quadratic formula
The quadratic formula is given by:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 8 \), \( b = -6 \), and \( c = 1 \).

### Step 3: Calculate the discriminant
First, we calculate the discriminant \( D \):
\[ D = b^2 - 4ac = (-6)^2 - 4 \cdot 8 \cdot 1 = 36 - 32 = 4 \]

### Step 4: Substitute values into the quadratic formula
Now substitute \( D \) back into the quadratic formula:
\[ y = \frac{-(-6) \pm \sqrt{4}}{2 \cdot 8} = \frac{6 \pm 2}{16} \]

### Step 5: Solve for \( y \)
This gives us two possible values for \( y \):
1. \( y = \frac{6 + 2}{16} = \frac{8}{16} = \frac{1}{2} \)
2. \( y = \frac{6 - 2}{16} = \frac{4}{16} = \frac{1}{4} \)

### Step 6: Convert back to \( \sec \theta \)
Now we have:
\[ \sec \theta = \frac{1}{2} \quad \text{or} \quad \sec \theta = \frac{1}{4} \]

### Step 7: Find the corresponding \( \cos \theta \)
Since \( \sec \theta = \frac{1}{\cos \theta} \), we can find \( \cos \theta \):
1. From \( \sec \theta = \frac{1}{2} \), we get \( \cos \theta = 2 \)
2. From \( \sec \theta = \frac{1}{4} \), we get \( \cos \theta = 4 \)

### Step 8: Analyze the values of \( \cos \theta \)
The cosine function has a range of \([-1, 1]\). Both values \( 2 \) and \( 4 \) are outside this range, meaning they are not possible.

### Conclusion
Since both possible values for \( \cos \theta \) are invalid, the original quadratic equation has **no roots**.

### Final Answer
The quadratic equation \( 8 \sec^2 \theta - 6 \sec \theta + 1 = 0 \) has **no roots**.

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