Question

x^2-(√2+1)x+√2=0
solve the equation by completing the square
​

Answer 1

Answer:

${x}^{2} - ( \sqrt{2} + 1)x + \sqrt{2} = 0$

${x}^{2} - 2x( \frac{ \sqrt{2} + 1}{2} ) + (\frac{ \sqrt{2} + 1}{2}) {}^{2} + \sqrt{2} - (\frac{ \sqrt{2} + 1}{2}) {}^{2} = 0$

$(x - \frac{ \sqrt{2} + 1}{2}) {}^{2} + \sqrt{2} - \frac{3 + 2 \sqrt{2} }{4} = 0$

$(x - \frac{ \sqrt{2} + 1}{2}) {}^{2} + \frac{4 \sqrt{2} - 3 - 2 \sqrt{2} }{4} = 0$

$(x - \frac{ \sqrt{2} + 1}{2}) {}^{2} = \frac{3 - 2 \sqrt{2} }{4} = ( \frac{ \sqrt{2} - 1 }{2} ) {}^{2}$

$x - \frac{ \sqrt{2} + 1}{2} = \frac{ \sqrt{2} - 1}{2} \: or \: - \frac{ \sqrt{2} - 1}{2}$

$x = \frac{ \sqrt{2} + 1}{2} + \frac{ \sqrt{2} - 1}{2} \: or \: \frac{ \sqrt{2} + 1}{2} - \frac{ \sqrt{2} - 1}{2}$

$x = \sqrt{2} \: \: or \: \: 1$

Ans.