Question

Y^2+1/2y-1=0 solve by completing square

Answer 1

SOLUTION:-

Completing square method;

${y}^{2} + \frac{1}{2} y - 1 = 0 \\ \\ = > {y}^{2} + \frac{1}{2} y = 1 \\ \\ = > {y}^{2} + 2 \times \frac{1}{4} y = 1 \\ \\ = > {y}^{2} + 2 \times \frac{1}{4} y + ( \frac{1}{4} ) {}^{2} = 1 + ( \frac{1}{4} ) {}^{2} \\ \\ = > (y + \frac{1}{4} ) {}^{2} = 1 + \frac{1}{16} \\ \\ = > (y + \frac{1}{4} ) {}^{2} = \frac{16 + 1}{16} \\ \\ = > (y + \frac{1}{4} ) {}^{2} = \frac{17}{16} \\ \\ = > (y + \frac{1}{4}) = \frac{ \sqrt{17} }{4} \\ \\ = > (y + \frac{1}{4} ) = \frac{ \sqrt{17} }{4} \: \: \: and \: \: (y + \frac{1}{4} ) = - \frac{ \sqrt{17} }{4} \\ \\ = > y = \frac{ \sqrt{17} }{4} - \frac{1}{4} \: \: \: \: and \: \: \: y = - \frac{ \sqrt{17} }{4} - \frac{1}{4} \\ \\ = > y = \frac{ \sqrt{17} - 1 }{4} \: \: \: and \: \: \: y = -\frac{ \sqrt{17} + 1 }{4}$

Hope it helps ☺️