Quadratic Equations
Location of roots
Grade 11
Question:
<p><strong>327.</strong> The smallest positive integral value of <em>a</em> for which the greater root of the equation \(x^2 - (a^2 + a + 1)x + a(a^2 + 1) = 0\) lies between the roots of the equation \(x^2 - a^2x - 2(a^2 - 2) = 0\), is less than:</p>
<p>\(\sqrt{\dfrac{27}{\sqrt{\dfrac{27}{\sqrt{\dfrac{27}{\sqrt{\cdots}}}}}}}\)</p>
<p>\(\sqrt{4\sqrt{4\sqrt{4\sqrt{\cdots}}}}\)</p>
<p>\(\sqrt{5}\sqrt[4]{5}\sqrt[8]{5}\sqrt[16]{5}\cdots\)</p>
<p>\(\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{\cdots}}}}\)</p>
Step-by-Step Solution
Key Concept: For the greater root of the first equation to lie between the roots of the second equation, the greater root must satisfy f(x) < 0 when substituted into the second quadratic. Factor the first equation to find its roots explicitly, then apply the interlacing condition.
<p><strong>Step 1: Factor the first equation.</strong></p><p>x² - (a² + a + 1)x + a(a² + 1) = 0</p><p>Testing factorization: (x - a)(x - (a² + 1)) = x² - (a² + 1)x - ax + a(a² + 1) = x² - (a² + a + 1)x + a(a² + 1) ✓</p><p>Roots are: x = a and x = a² + 1</p><p><strong>Step 2: Identify the greater root.</strong></p><p>For positive a: a² + 1 > a, so the greater root is α = a² + 1</p><p><strong>Step 3: Find roots of the second equation.</strong></p><p>x² - a²x - 2(a² - 2) = 0</p><p>Let β₁ and β₂ be its roots with β₁ < β₂</p><p><strong>Step 4: Apply the interlacing condition.</strong></p><p>For α to lie between β₁ and β₂: g(α) < 0, where g(x) = x² - a²x - 2(a² - 2)</p><p>g(a² + 1) = (a² + 1)² - a²(a² + 1) - 2(a² - 2)</p><p>= a⁴ + 2a² + 1 - a⁴ - a² - 2a² + 4</p><p>= (2a² - a² - 2a²) + 1 + 4</p><p>= -a² + 5 < 0</p><p><strong>Step 5: Solve for a.</strong></p><p>-a² + 5 < 0</p><p>a² > 5</p><p>a > √5 ≈ 2.236</p><p>The smallest positive integer satisfying this is a = 3</p><p>Therefore, the answer is less than 4.</p><p>∴ Answer: B</p>
Correct Answer: B