Paper 2 · Errors in proofs · Algebra and functions
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A student is asked to find all real solutions of x+6=x\sqrt{x + 6} = x. Here is their argument. (I) x+6=x\sqrt{x + 6} = x (II) Squaring both sides, x+6=x2x + 6 = x^2. (III) So x2−x−6=0x^2 - x - 6 = 0. (IV) So (x−3)(x+2)=0(x - 3)(x + 2) = 0. (V) Therefore the solutions are x=3x = 3 and x=−2x = -2. Which of the following is true?