Mathematics · Quantitative Aptitude

Equations and Roots

65 Questions

Equations and roots questions cover finding real solutions to algebraic, trigonometric, and polynomial equations. They are fundamental for advanced mathematics sections. Practicing these problems ensures quick recognition of underlying patterns and calculation shortcuts.

polynomial equationssystem of equationstrigonometric equationsabsolute value equations

Equations and Roots Questions

Multiple choice

Find all solutions of the equation (\sec^2\theta - 2\sec\theta - 3 = 0) in the interval ([0, 2\pi)).

  1. \(\theta = \frac{\pi}{3}, \frac{5\pi}{3}\)
  2. \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)
  3. \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)
  4. \(\theta = 0, \pi\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the identity (\sec^2\theta = 1 + \tan^2\theta), we can rewrite the equation as (1 + \tan^2\theta - 2\tan\theta - 3 = 0). Expanding and rearranging, we get (\tan^2\theta - 2\tan\theta - 4 = 0). Factoring, we find ((\tan\theta - 4)(\tan\theta + 1) = 0). Solving each factor separately, we find (\tan\theta = 4) or (\tan\theta = -1). Using the unit circle or reference angles, we find the solutions (\theta = \frac{\pi}{3}, \frac{5\pi}{3}).

Multiple choice

Find all solutions of the equation (2\sin^2\theta - 3\sin\theta + 1 = 0) in the interval ([0, 2\pi)).

  1. \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)
  2. \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}\)
  3. \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)
  4. \(\theta = 0, \pi\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Factoring the equation, we get ((2\sin\theta - 1)(\sin\theta - 1) = 0). Solving each factor separately, we find (\sin\theta = \frac{1}{2}) or (\sin\theta = 1). Using the unit circle or reference angles, we find the solutions (\theta = \frac{\pi}{6}, \frac{5\pi}{6}).

Multiple choice

Find all solutions of the equation (\cot^2\theta - 3\cot\theta + 2 = 0) in the interval ([0, 2\pi)).

  1. \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)
  2. \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}\)
  3. \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)
  4. \(\theta = 0, \pi\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Factoring the equation, we get ((\cot\theta - 2)(\cot\theta - 1) = 0). Solving each factor separately, we find (\cot\theta = 2) or (\cot\theta = 1). Using the unit circle or reference angles, we find the solutions (\theta = \frac{\pi}{6}, \frac{5\pi}{6}).

Multiple choice

Find all solutions of the equation (\csc^2\theta - 2\csc\theta - 3 = 0) in the interval ([0, 2\pi)).

  1. \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}\)
  2. \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}\)
  3. \(\theta = \frac{\pi}{4}, \frac{3\pi}{4}\)
  4. \(\theta = 0, \pi\)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the identity (\csc^2\theta = 1 + \cot^2\theta), we can rewrite the equation as (1 + \cot^2\theta - 2\csc\theta - 3 = 0). Expanding and rearranging, we get (\cot^2\theta - 2\csc\theta - 2 = 0). Factoring, we find ((\cot\theta - 2)(\cot\theta + 1) = 0). Solving each factor separately, we find (\cot\theta = 2) or (\cot\theta = -1). Using the unit circle or reference angles, we find the solutions (\theta = \frac{\pi}{6}, \frac{5\pi}{6}).

Multiple choice

Find the number of solutions to the equation x^2 + 2x + 1 = 0 in the field of complex numbers.

  1. 0

  2. 1

  3. 2

  4. 3
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation x^2 + 2x + 1 = 0 is a quadratic equation. Using the quadratic formula, we have: x = (-2 ± √(2^2 - 4 * 1 * 1)) / (2 * 1) = -1 ± √3i. Therefore, there are two solutions to the equation in the field of complex numbers.