Question

Use completing the square method using K
9y square - 12y +2 = 0

Answer 1

$9 {y}^{2} - 12y + 2 = 0 \\ \\ \Rightarrow \frac{9 {y}^{2} - 12y + 2}{9} = \frac{0}{9} \\ \\ \Rightarrow {y }^{2} - \frac{12}{9} \: y + \frac{2}{9} = 0 \\ \\ \Rightarrow {y }^{2} - \frac{12}{9} \: y = - \frac{2}{9}\\ \\ \Rightarrow {(y )}^{2} - 2 \times (\frac{6}{9}) \times y + {( \frac{6}{9}) }^{2} = - \frac{2}{9} + {( \frac{6}{9}) }^{2} \\ \\ \Rightarrow {(y - \frac{6}{9}) }^{2} = \frac{36}{81} - \frac{2}{9} \\ \\ \Rightarrow {(y - \frac{6}{9}) }^{2} = \frac{36 - 18}{81} = \frac{18}{81} = \frac{2}{9} \\ \\ \Rightarrow {(y - \frac{2}{3}) }^{2} = \frac{2}{9} \\ \\ \Rightarrow y - \frac{2}{3} = \pm \frac{ \sqrt{2} }{3} \\ \\ \Rightarrow y = \frac{2}{3} \pm \frac{ \sqrt{2} }{3} \\ \\ \Rightarrow \boxed{y = \frac{2 + \sqrt{2} }{3} \: \: and \: \: \frac{2 - \sqrt{2} }{3}}$

Answer 2

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$9 {y}^{2} - 12y + 2 = 0 \\ \\ \Rightarrow \frac{9 {y}^{2} - 12y + 2}{9} = \frac{0}{9} \\ \\ \Rightarrow {y }^{2} - \frac{12}{9} \: y + \frac{2}{9} = 0 \\ \\ \Rightarrow {y }^{2} - \frac{12}{9} \: y = - \frac{2}{9}\\ \\ \Rightarrow {(y )}^{2} - 2 \times (\frac{6}{9}) \times y + {( \frac{6}{9}) }^{2} = - \frac{2}{9} + {( \frac{6}{9}) }^{2} \\ \\ \Rightarrow {(y - \frac{6}{9}) }^{2} = \frac{36}{81} - \frac{2}{9} \\ \\ \Rightarrow {(y - \frac{6}{9}) }^{2} = \frac{36 - 18}{81} = \frac{18}{81} = \frac{2}{9} \\ \\ \Rightarrow {(y - \frac{2}{3}) }^{2} = \frac{2}{9} \\ \\ \Rightarrow y - \frac{2}{3} = \pm \frac{ \sqrt{2} }{3} \\ \\ \Rightarrow y = \frac{2}{3} \pm \frac{ \sqrt{2} }{3} \\ \\ \Rightarrow \boxed{y = \frac{2 + \sqrt{2} }{3} \: \: and \: \: \frac{2 - \sqrt{2} }{3}}$

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