Quadratic Equations
Vieta's Formulas
Grade 11
Question:
<p>Let $-\frac{\pi}{6} \leq I \leq \frac{\pi}{12}$. Suppose $r_1$ and $s_1$ are the roots of equation $x^2 - 2x\sec I + 1 = 0$ and $r_2$ and $s_2$ are the roots of equation $x^2 + 2x\tan I - 1 = 0$. If $r_1 > s_1$ and $r_2 > s_2$, then $r_1 + s_2$ equals</p>
<p>(a) $2(\sec I - \tan I)$</p>
<p>(b) $2\sec I$</p>
<p>(c) $-2\tan I$</p>
<p>(d) $0$</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to find the roots in terms of trigonometric functions, then carefully determine the signs based on the given range of I.
<p><strong>Solution:</strong> For the equation $x^2 - 2x\sec I + 1 = 0$:</p><p>By Vieta's formulas: $r_1 + s_1 = 2\sec I$ and $r_1 \cdot s_1 = 1$</p><p>Since $r_1 > s_1$: $r_1 = \sec I + \sqrt{\sec^2 I - 1} = \sec I + |\tan I|$</p><p>For the equation $x^2 + 2x\tan I - 1 = 0$:</p><p>By Vieta's formulas: $r_2 + s_2 = -2\tan I$ and $r_2 \cdot s_2 = -1$</p><p>Since $r_2 > s_2$: $r_2 = -\tan I + \sqrt{\tan^2 I + 1} = -\tan I + |\sec I|$</p><p>Given $-\frac{\pi}{6} \leq I \leq \frac{\pi}{12}$, we have $\sec I > 0$ and $\tan I \leq 0$, so $|\tan I| = -\tan I$ and $|\sec I| = \sec I$.</p><p>Therefore: $r_1 = \sec I - \tan I$ and $s_2 = -\tan I - \sec I$</p><p>Thus: $r_1 + s_2 = (\sec I - \tan I) + (-\tan I - \sec I) = 2(\sec I - \tan I)$</p><p>∴ Answer is (a).</p>
Correct Answer: A