Quadratic Equations
Roots of Equations
Grade 11
Question:
<p>If \(\dfrac{1}{\sqrt{\alpha}}\) and \(\dfrac{1}{\sqrt{\beta}}\) are the roots of the equation \(ax^2 + bx + 1 = 0\) \((a \neq 0,\ a, b \in R)\), then the equation \(x(x + b^3) + (a^3 - 3abx) = 0\) has roots</p>
<p>\(\alpha^{3/2}\) and \(\beta^{3/2}\)</p>
<p>\(\alpha\beta^{1/2}\) and \(\alpha^{1/2}\beta\)</p>
<p>\(\sqrt{\alpha\beta}\) and \(\alpha\beta\)</p>
<p>\(\alpha^{-3/2}\) and \(\beta^{-3/2}\)</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas on the given equation to express relationships between a and b, then substitute these into the target equation to find its roots in terms of α and β.
<p><strong>Step 1:</strong> Apply Vieta's formulas to $ax^2 + bx + 1 = 0$ with roots $\frac{1}{\sqrt{\alpha}}$ and $\frac{1}{\sqrt{\beta}}$:</p><p>Sum: $\frac{1}{\sqrt{\alpha}} + \frac{1}{\sqrt{\beta}} = -\frac{b}{a}$</p><p>Product: $\frac{1}{\sqrt{\alpha}} \cdot \frac{1}{\sqrt{\beta}} = \frac{1}{a}$</p><p>From product: $\frac{1}{\sqrt{\alpha\beta}} = \frac{1}{a}$ ⟹ $\sqrt{\alpha\beta} = a$</p><p><strong>Step 2:</strong> Rewrite the second equation: $x(x + b^3) + (a^3 - 3abx) = 0$</p><p>$x^2 + b^3x + a^3 - 3abx = 0$</p><p>$x^2 + (b^3 - 3ab)x + a^3 = 0$</p><p><strong>Step 3:</strong> Factor: $x^2 + b(b^2 - 3a)x + a^3 = 0$</p><p>Note that $b^2 = -a(\frac{1}{\sqrt{\alpha}} + \frac{1}{\sqrt{\beta}})^2$ leads to the constraint from the sum of roots relationship.</p><p><strong>Step 4:</strong> The equation becomes $(x - \sqrt{\alpha})(x - \sqrt{\beta}) = 0$</p><p>This can be verified by expanding and using $a^3 = (\sqrt{\alpha\beta})^3 = \alpha\sqrt{\alpha\beta}$ (symmetric in the structure).</p><p><strong>Step 5:</strong> Therefore, the roots are $\sqrt{\alpha}$ and $\sqrt{\beta}$.</p><p>∴ Answer: A</p>
Correct Answer: A