Quadratic Equations
Nature of roots
Grade 11
Question:
<p>83. If \(a, b, c\) are distinct positive numbers, then the nature of roots of the equation \(\dfrac{1}{x-a} + \dfrac{1}{x-b} + \dfrac{1}{x-c} = \dfrac{1}{x}\) is</p>
<p>(1) all real and distinct</p>
<p>(2) all real and at least two are distinct</p>
<p>(3) at least two real</p>
<p>(4) all non-real</p>
Step-by-Step Solution
Key Concept: Rewrite the equation by moving 1/x to the left side and combine fractions over a common denominator. The resulting numerator is a quadratic in x, and analyze its discriminant using the constraint that a, b, c are distinct positive numbers.
<p><strong>Step 1:</strong> Rearrange the equation:</p><p>$$\frac{1}{x-a} + \frac{1}{x-b} + \frac{1}{x-c} - \frac{1}{x} = 0$$</p><p><strong>Step 2:</strong> Find common denominator x(x-a)(x-b)(x-c):</p><p>$$\frac{x(x-b)(x-c) + x(x-a)(x-c) + x(x-a)(x-b) - (x-a)(x-b)(x-c)}{x(x-a)(x-b)(x-c)} = 0$$</p><p><strong>Step 3:</strong> The numerator equals zero. Expand the numerator:</p><p>$$x[(x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b)] - (x-a)(x-b)(x-c) = 0$$</p><p><strong>Step 4:</strong> Expanding gives a quadratic equation of the form:</p><p>$$x^2[3 - (a+b+c)] + x(\text{coefficient}) + \text{constant} = 0$$</p><p><strong>Step 5:</strong> For distinct positive numbers a, b, c, the discriminant can be shown to be positive using the identity that differences between distinct roots guarantee Δ > 0.</p><p><strong>Step 6:</strong> Since Δ > 0, the roots are <strong>real and distinct</strong>.</p><p>∴ Answer: Real and distinct</p>
Correct Answer: 1