For which values of with \;0 < < /2\; does the quadratic in t given by \;t^2 + 4 t + \; have repeated roots?

Quadratics Trigonometric Equations Difficulty 4

TMUA practice: Quadratics, problem 4

A TMUA-calibre quadratics problem at difficulty 4 of 5, with the full worked solution.

The question

For which values of

\theta

with

\;0 < \theta < \pi/2\;

does the quadratic in

t

given by

\;t^{2} + 4 t \cos\theta + \cot\theta\;

have repeated roots?

\tfrac{\pi}{6} \text{ or } \tfrac{5\pi}{18}

\tfrac{\pi}{6} \text{ or } \tfrac{5\pi}{12}

\tfrac{\pi}{12} \text{ or } \tfrac{5\pi}{18}

\tfrac{\pi}{12} \text{ or } \tfrac{5\pi}{12}

The correct answer is highlighted. D

Worked solution

A quadratic $\;a t^{2} + b t + c\;$ has repeated roots iff its discriminant is zero. Here $a = 1$ , $b = 4\cos\theta$ , $c = \cot\theta$ , so

\Delta \;=\; (4 \cos\theta)^{2} - 4(1)(\cot\theta) \;=\; 16 \cos^{2}\theta - 4 \cot\theta.

Set $\Delta = 0$ and use $\cot\theta = \cos\theta / \sin\theta$ :

$16 \cos^{2}\theta \;=\; \frac{4 \cos\theta}{\sin\theta}.$

For $\theta \in (0, \pi/2)$ we have $\cos\theta \ne 0$ , so divide by $4\cos\theta$ :

4 \cos\theta \;=\; \frac{1}{\sin\theta} \;\Longleftrightarrow\; 4 \sin\theta \cos\theta \;=\; 1 \;\Longleftrightarrow\; 2 \sin(2\theta) \;=\; 1 \;\Longleftrightarrow\; \sin(2\theta) \;=\; \tfrac{1}{2}.

Solutions in $\;2\theta \in (0, \pi)$ : $\;2\theta = \pi/6\;$ or $\;2\theta = 5\pi/6$ . Hence

$\theta \;=\; \tfrac{\pi}{12} \quad\text{or}\quad \theta \;=\; \tfrac{5\pi}{12}.$

This is option D.

Why the other options fail. A, B, C all replace one or both of these values incorrectly (e.g. $5\pi/18$ would arise from $2\theta = 5\pi/9$ , which is not a solution of $\sin(2\theta) = 1/2$ ).

The lesson: ‘repeated roots’ is the discriminant-zero condition. Combine $\cot\theta = \cos\theta / \sin\theta$ and the double-angle identity $\;2 \sin\theta \cos\theta = \sin(2\theta)$ .