<p>If \(a \in (-1, 1)\), then roots of the quadratic equation \((a-1)x^2 + ax + \sqrt{1 - a^2} = 0\) are</p>
Step-by-Step Solution
Key Concept: When $a \in (-1,1)$, we have $a-1 < 0$ and $\sqrt{1-a^2}$ is real. The discriminant $\Delta = a^2 - 4(a-1)\sqrt{1-a^2}$ must be analyzed carefully by recognizing that $(a-1)\sqrt{1-a^2} < 0$, making $\Delta > 0$, ensuring real distinct roots.
<p><strong>Step 1:</strong> Identify the quadratic equation $(a-1)x^2 + ax + \sqrt{1-a^2} = 0$ where $a \in (-1,1)$.</p><p><strong>Step 2:</strong> Note that $a-1 < 0$ (so equation is quadratic) and $\sqrt{1-a^2}$ is real since $1-a^2 > 0$.</p><p><strong>Step 3:</strong> Calculate discriminant: $\Delta = a^2 - 4(a-1)\sqrt{1-a^2}$. Since $a-1 < 0$, we have $-4(a-1)\sqrt{1-a^2} > 0$, so $\Delta > a^2 > 0$.</p><p><strong>Step 4:</strong> Use the quadratic formula: $x = \frac{-a \pm \sqrt{a^2 - 4(a-1)\sqrt{1-a^2}}}{2(a-1)}$. The expression under the square root can be rewritten as $a^2 + 4(1-a)\sqrt{1-a^2} = [a + 2\sqrt{1-a^2}]^2 - 4(1-a^2) + 4(1-a)\sqrt{1-a^2}$, which simplifies to yield real distinct roots.</p><p><strong>Step 5:</strong> The roots are real and distinct (since $\Delta > 0$) for all $a \in (-1,1)$.</p><p>∴ Answer: B</p>
Correct Answer: B