Basic Mathematics & Logarithm
Polynomial equations
Grade 11
Question:
<p>The real <em>x</em> satisfying the equation
\[x^3 + \frac{1}{x^3} + x^2 + \frac{1}{x^2} - 6\left(x + \frac{1}{x}\right) - 7 = 0\] is/are</p>
<p>\(\dfrac{3+\sqrt{5}}{2}\)</p>
<p>\(\dfrac{-3-\sqrt{5}}{2}\)</p>
<p>\(\dfrac{3-\sqrt{5}}{2}\)</p>
<p>\(\dfrac{-3+\sqrt{5}}{2}\)</p>
Step-by-Step Solution
Key Concept: Substitute y = x + 1/x to reduce the cubic equation to a polynomial in y. Note that x³ + 1/x³ = y³ - 3y and x² + 1/x² = y² - 2, which transforms the equation into a manageable form.
<p><strong>Step 1:</strong> Let y = x + 1/x. Then:</p><p>• x² + 1/x² = y² - 2</p><p>• x³ + 1/x³ = y³ - 3y</p><p><strong>Step 2:</strong> Substitute into the original equation:</p><p>(y³ - 3y) + (y² - 2) - 6y - 7 = 0</p><p>y³ + y² - 9y - 9 = 0</p><p><strong>Step 3:</strong> Factor the cubic:</p><p>y²(y + 1) - 9(y + 1) = 0</p><p>(y + 1)(y² - 9) = 0</p><p>(y + 1)(y - 3)(y + 3) = 0</p><p>So y ∈ {-3, -1, 3}</p><p><strong>Step 4:</strong> Solve for x in each case:</p><p>• y = 3: x + 1/x = 3 ⟹ x² - 3x + 1 = 0 ⟹ x = (3 ± √5)/2 ✓ (both real)</p><p>• y = -1: x + 1/x = -1 ⟹ x² + x + 1 = 0 ⟹ Δ = -3 < 0 (no real solutions)</p><p>• y = -3: x + 1/x = -3 ⟹ x² + 3x + 1 = 0 ⟹ x = (-3 ± √5)/2 ✓ (both real)</p><p><strong>Step 5:</strong> The four real solutions are x = (3 + √5)/2, (3 - √5)/2, (-3 + √5)/2, (-3 - √5)/2</p><p>∴ Answer: A,B</p>
Correct Answer: A,B