Quadratic Equations
Nature of roots
Grade 11
Question:
<p>If one root of \(x^2 - x - k = 0\) is square of the other, then \(k =\)</p>
<p>\(2 \pm \sqrt{5}\)</p>
<p>\(2 \pm \sqrt{3}\)</p>
<p>\(3 \pm \sqrt{2}\)</p>
<p>\(5 \pm \sqrt{2}\)</p>
Step-by-Step Solution
Key Concept: If one root is the square of the other, use Vieta's formulas combined with the constraint that if α is a root, then α² is also a root. This creates a system: α + α² = 1 (sum) and α · α² = -k (product).
<p><strong>Step 1:</strong> Let the roots be α and α². By Vieta's formulas for x² - x - k = 0:</p><ul><li>Sum of roots: α + α² = 1</li><li>Product of roots: α · α² = -k, so α³ = -k</li></ul><p><strong>Step 2:</strong> From α + α² = 1, we get α² = 1 - α. Substitute into the quadratic:</p><p>α² - α - k = 0</p><p>(1 - α) - α - k = 0</p><p>1 - 2α - k = 0 ... (i)</p><p><strong>Step 3:</strong> Since α is a root: α² - α - k = 0, and α² = 1 - α, we have:</p><p>(1 - α) - α - k = 0 ⟹ k = 1 - 2α</p><p><strong>Step 4:</strong> From α + α² = 1 and α³ = -k, we have α² = 1 - α. So:</p><p>α³ = α(1 - α) = α - α² = α - (1 - α) = 2α - 1</p><p>Thus: -k = 2α - 1 ⟹ k = 1 - 2α</p><p><strong>Step 5:</strong> Solving α² + α - 1 = 0 (from α + α² = 1 rearranged):</p><p>α = (-1 ± √5)/2</p><p>For α = (-1 + √5)/2: k = 1 - 2(-1 + √5)/2 = 1 + 1 - √5 = <strong>2 - √5</strong></p><p>For α = (-1 - √5)/2: k = 1 - 2(-1 - √5)/2 = 1 + 1 + √5 = <strong>2 + √5</strong></p><p>∴ Answer: A (typically k = 2 or check given options for k = 2 ± √5)</p>
Correct Answer: A