Multiple choice

The number of roots of the quadratic equation $8{\sec ^2}\theta - 6\sec \theta + 1 = 0$ is:

  1. Infinite
  2. $1$
  3. $2$
  4. $0$
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D Correct answer
Explanation

Let x = sec(theta). The equation is 8x^2 - 6x + 1 = 0. Factoring gives (4x - 1)(2x - 1) = 0, so x = 1/4 or x = 1/2. Since the range of sec(theta) is (-inf, -1] U [1, inf), neither 1/4 nor 1/2 are possible values for sec(theta). Thus, there are no real roots.

AI explanation

Treating $\sec \theta$ as a variable $y$, we solve the quadratic equation $8y^2 - 6y + 1 = 0$. Factoring the quadratic gives $(4y - 1)(2y - 1) = 0$, revealing the roots $y = 1/4$ and $y = 1/2$. This means $\sec \theta = 1/4$ or $\sec \theta = 1/2$, which further implies $\cos \theta = 4$ or $\cos \theta = 2$. Since the range of the cosine function is restricted to $[-1, 1]$, neither of these values is possible, so the equation has 0 roots.

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