SSC CHSL 13 Nov 2025 Shift 3 Paper

Concept:This is a rational equation. Combine the fractions on the left into a single fraction and then solve the resulting quadratic equation.Explanation:Start with $\frac{x}{x+1} + \frac{1}{x} = 3$.Combine the fractions: $\frac{x \cdot x + 1 \cdot (x+1)}{x(x+1)} = 3$.This gives $\frac{x^2 + x + 1}{x(x+1)} = 3$.Multiply both sides by $x(x+1)$: $x^2 + x + 1 = 3x(x+1)$.Expand the right side: $x^2 + x + 1 = 3x^2 + 3x$.Bring all terms to one side: $0 = 3x^2 + 3x - x^2 - x - 1$ → $0 = 2x^2 + 2x - 1$.So $2x^2 + 2x - 1 = 0$.Apply the quadratic formula: $x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}$.Simplify: $x = \frac{-2 \pm \sqrt{4 + 8}}{4} = \frac{-2 \pm \sqrt{12}}{4} = \frac{-2 \pm 2\sqrt{3}}{4}$.Divide numerator and denominator by 2: $x = \frac{-1 \pm \sqrt{3}}{2}$.Thus the solutions are $x = \frac{-1 + \sqrt{3}}{2}$ or $x = \frac{-1 - \sqrt{3}}{2}$.Answer:Option B: $x = \frac{-1 + \sqrt{3}}{2}$ is one of the correct roots.