Quadratic Equations
Roots of Equations
Grade 11
Question:
<p>Let \(\alpha\) and \(\beta\) be the roots of equation \(x^2 - 6x - 2 = 0\). If \(a_n = \alpha^n - \beta^n\), for \(n \geq 1\), then the value of \(\dfrac{a_{10} - 2a_8}{2a_9}\) is equal to</p>
<p>\(-6\)</p>
<p>\(3\)</p>
<p>\(-3\)</p>
<p>\(6\)</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to establish a recurrence relation for $a_n = \alpha^n - \beta^n$, then derive that $a_n = 6a_{n-1} + 2a_{n-2}$ from the characteristic equation property.
<p><strong>Step 1:</strong> From $x^2 - 6x - 2 = 0$, by Vieta's formulas: $\alpha + \beta = 6$ and $\alpha\beta = -2$</p><p><strong>Step 2:</strong> Since $\alpha$ and $\beta$ are roots: $\alpha^2 = 6\alpha + 2$ and $\beta^2 = 6\beta + 2$</p><p><strong>Step 3:</strong> For $a_n = \alpha^n - \beta^n$, multiply the recurrence by $\alpha^{n-2}$ and $\beta^{n-2}$:<br/>$\alpha^n = 6\alpha^{n-1} + 2\alpha^{n-2}$ and $\beta^n = 6\beta^{n-1} + 2\beta^{n-2}$</p><p><strong>Step 4:</strong> Subtracting: $a_n = 6a_{n-1} + 2a_{n-2}$ for all $n \geq 2$</p><p><strong>Step 5:</strong> Therefore: $a_{10} = 6a_9 + 2a_8$</p><p><strong>Step 6:</strong> Substitute into the expression:<br/>$$\frac{a_{10} - 2a_8}{2a_9} = \frac{(6a_9 + 2a_8) - 2a_8}{2a_9} = \frac{6a_9}{2a_9} = 3$$</p><p>∴ Answer: B</p>
Correct Answer: B