Quadratic Equations
Location of roots
Grade 11
Question:
<p>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>\(\sqrt{4\sqrt{4\sqrt{4\cdots}}}\)</p>
<p>\(\sqrt{5}\sqrt[4]{5}\sqrt[8]{5}\sqrt[16]{5}\cdots\)</p>
<p>\(\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}\)</p>
Step-by-Step Solution
Key Concept: For a root of the first equation to lie between the roots of the second equation, the first equation's root must satisfy the second equation with opposite sign inequality. Use the property that if α lies between roots of f(x)=0, then f(α) has opposite sign to the leading coefficient.
<p><strong>Step 1:</strong> Factor the first equation: x² - (a² + a + 1)x + a(a² + 1) = 0</p><p>Factoring: (x - a)(x - (a² + 1)) = 0</p><p>Roots are x = a and x = a² + 1. Since a > 0, the greater root is α = a² + 1.</p><p><strong>Step 2:</strong> For α = a² + 1 to lie between the roots of x² - a²x - 2(a² - 2) = 0, we need f(α) < 0 where f(x) = x² - a²x - 2(a² - 2).</p><p><strong>Step 3:</strong> Substitute α = a² + 1 into f(x):</p><p>f(a² + 1) = (a² + 1)² - a²(a² + 1) - 2(a² - 2)</p><p>= a⁴ + 2a² + 1 - a⁴ - a² - 2a² + 4</p><p>= a² + 1 - a² - 2a² + 4</p><p>= -2a² + 5</p><p><strong>Step 4:</strong> We need f(a² + 1) < 0:</p><p>-2a² + 5 < 0</p><p>5 < 2a²</p><p>a² > 5/2 = 2.5</p><p>a > √2.5 ≈ 1.58</p><p><strong>Step 5:</strong> The smallest positive integer satisfying a² > 2.5 is a = 2 (since 2² = 4 > 2.5, but 1² = 1 < 2.5).</p><p>∴ Answer: B</p>
Correct Answer: B