WAEC Further Mathematics 2024 Paper

Concept:Use substitution to transform the irrational equation into a quadratic equation, then reject any impossible root.Explanation:Start with the given equation:$$\sqrt{x} - \frac{6}{\sqrt{x}} = 1$$Multiply every term by $\sqrt{x}$ to remove the denominator:$$x - 6 = \sqrt{x}$$Rearrange the equation:$$x - \sqrt{x} - 6 = 0$$Let $q = \sqrt{x}$. Then $x = q^2$. Substitute into the equation:$$q^2 - q - 6 = 0$$Factorise the quadratic:$$q^2 - 3q + 2q - 6 = 0$$$$q(q - 3) + 2(q - 3) = 0$$$$(q - 3)(q + 2) = 0$$So, $q = 3$ or $q = -2$.Since $q = \sqrt{x}$ must be non-negative, reject $q = -2$.Therefore, $q = 3$, which gives:$$\sqrt{x} = 3$$Squaring both sides:$$x = 9$$Verify: $\sqrt{9} - \frac{6}{\sqrt{9}} = 3 - 2 = 1$, which is correct.Answer:$$x = 9$$This corresponds to option C.