WAEC Further Mathematics 2013 Paper

Concept:Squaring a radical equation removes the square roots and gives a simpler equation to solve.Explanation:Start with $\sqrt{x} + \sqrt{x+1} = \sqrt{2x+1}$.Square both sides:$\left(\sqrt{x} + \sqrt{x+1}\right)^{2} = \left(\sqrt{2x+1}\right)^{2}$.Expand using $(a+b)^2 = a^2 + 2ab + b^2$:$x + 2\sqrt{x(x+1)} + x + 1 = 2x + 1$.Simplify the left side:$2x + 1 + 2\sqrt{x(x+1)} = 2x + 1$.Subtract $2x + 1$ from both sides:$2\sqrt{x(x+1)} = 0$.So $\sqrt{x(x+1)} = 0$.Square both sides again:$x(x+1) = 0$.Hence $x = 0$ or $x = -1$.Both values satisfy the original equation, with $x=-1$ giving $\sqrt{-1}$ in the complex form $i$.Answer:The possible values of $x$ are $0$ and $-1$.