Question 27: Find the value of x and y such that \({\left( {x – 3} \right)^2} + \sqrt {y – 2} = 0\)

We have: \({\left( {x – 3} \right)^2} \ge 0\) and \(\sqrt {y – 2} \ge 0.\) Hence \({\left( { {) x – 3} \right)^2} + \sqrt {y – 2} = 0\)

There should be: \({\left( {x – 3} \right)^2} = 0\) and \(\sqrt {y – 2} = 0\)

\( \Rightarrow x – 3 = 0\) and \(y – 2 = 0 \Rightarrow x = 3;y = 2\)