Question

Solve the following quadratic equation by completing the square method.

9y² – 12y + 2 = 0

Answer 1

 9y² – 12y + 2 = 0 

\[\Rightarrow y^2 - \frac{12}{9}y + \frac{2}{9} = 0\]
\[ \Rightarrow y^2 - \frac{4}{3}y + \frac{2}{9} = 0\]
\[ \Rightarrow y^2 - \frac{4}{3}y + \left( \frac{\frac{4}{3}}{2} \right)^2 - \left( \frac{\frac{4}{3}}{2} \right)^2 + \frac{2}{9} = 0\]
\[ \Rightarrow y^2 - \frac{4}{3}y + \left( \frac{2}{3} \right)^2 - \left( \frac{2}{3} \right)^2 + \frac{2}{9} = 0\]
\[\Rightarrow \left[ y^2 - \frac{4}{3}y + \left( \frac{4}{9} \right) \right] - \left( \frac{4}{9} \right) + \frac{2}{9} = 0\]
\[ \Rightarrow \left( y - \frac{2}{3} \right)^2 - \frac{2}{9} = 0\]
\[ \Rightarrow \left( y - \frac{2}{3} \right)^2 = \frac{2}{9}\]
\[ \Rightarrow \left( y - \frac{2}{3} \right)^2 = \left( \frac{\sqrt{2}}{3} \right)^2 \]
\[\Rightarrow y - \frac{2}{3} = \frac{\sqrt{2}}{3} \text{ or } y - \frac{2}{3} = - \frac{\sqrt{2}}{3}\]
\[ \Rightarrow y = \frac{\sqrt{2}}{3} + \frac{2}{3} \text{ or } y = - \frac{\sqrt{2}}{3} + \frac{2}{3}\]
\[ \Rightarrow y = \frac{\sqrt{2} + 2}{3} \text{ or } y = \frac{- \sqrt{2} + 2}{3}\]