Question

The quadratic equation $\tan \theta x^2+2(\sec \theta +\cos \theta )x+(\tan \theta +3\sqrt{2}\cot \theta )$ always has

• A. equal roots for all $\theta$
• B. complex roots for all $\theta$
• C. real and distinct roots for all $\theta$
• D. real roots or complex roots depending upon $\theta$

Answer 1

The correct option is B

complex roots for all $\theta$

Discriminant, $△=b^2-4ac$
$=\left(2\right(\sec \theta +\cos \theta ){)}^2-4\tan \theta (\tan \theta +3\sqrt{2}\cot \theta )$
$=4({\sec }^2\theta +{\cos }^2\theta +2)-4({\tan }^2\theta +3\sqrt{2})]$
$=4\left[\right({\sec }^2\theta +{\cos }^2\theta +2-({\tan }^2\theta +3\sqrt{2})]$
$=4\left[\right({\sec }^2\theta +{\tan }^2\theta +2+{\cos }^2\theta +3\sqrt{2}\left)\right]$
$=4[1+2+{\cos }^2\theta -3\sqrt{2})$
$=4[3+{\cos }^2\theta -3\sqrt{2}]$
$=4[{\cos }^2\theta +3-3\sqrt{2}]$

As ${\cos }^2\theta \le 1$
So, $△\le 4[1+3-3\sqrt{2}]$
$=4(4-3\sqrt{2})$
$\simeq 4(4-3\times 1.4)$
$=4(-0.2)$
$\Rightarrow △$ is negative
$\Rightarrow$No real roots