Explaining a rejected root and a false surd rule

Explaining a rejected root and a false surd rule

Secondary 3

Do not use a calculator in this question.

  1. Amir solves \(\sqrt{2x+7}=x-4\). He squares both sides, obtains \(x^2-10x+9=0\) and gives two answers, \(x=1\) and \(x=9\). Explain why one of his answers must be rejected, and state the solution of the equation.
  2. Bea claims that \(\sqrt{x+9}=\sqrt{x}+3\) for every \(x\ge0\). Show that her claim is false, and find the only value of \(x\) for which it is true.

Solution:

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Answer:(a) \(x=1\) gives \(3=-3\), so it is rejected; \(x=9\) (b) false, e.g. \(x=16\) gives \(5\ne7\); true only for \(x=0\)

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