Question

Solve by using the method of completing the square :-

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

Answer 1

Hey there !!


▶ The given quadratic equation :-

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

$= > x {}^{2} - ( \sqrt{2} + 1)x = - \sqrt{2} . \\ \\ (adding {( \frac{ \sqrt{2} + 1 }{2}) }^{2} on \: both \: side ) \\ \\ = > {x}^{2} - 2 \times x \times ( \frac{ \sqrt{2} + 1}{2} ) + {( \frac{ \sqrt{2} + 1 }{2}) }^{2} = - \sqrt{2} + {( \frac{ \sqrt{2} + 1 }{2}) }^{2}. \\ \\ = > {[x - ( \frac{ \sqrt{2} + 1 }{2} )]}^{2} = - \sqrt{2} + \frac{2 + 1 + 2 \sqrt{2} }{4} . \\ \\ = > {[x - ( \frac{ \sqrt{2} + 1 }{2} )]}^{2} = \frac{ - 4 \sqrt{2} + 2 + 1 + 2 \sqrt{2} }{4} . \\ \\ = > {[x - ( \frac{ \sqrt{2} + 1 }{2} )]}^{2} = \frac{2 - 2 \sqrt{2} + 1}{4} . \\ \\ = > {[x - ( \frac{ \sqrt{2} + 1 }{2} )]}^{2} = {( \frac{ \sqrt{2} - 1}{2} )}^{2} . \\ \\ [ \: taking \: square \: root \: on \: both \: side ,we \: get \:] \\ \\ = > x - ( \frac{ \sqrt{2} + 1 }{2} ) = ±( \frac{ \sqrt{2} - 1 }{2} ). \\ \\ = > x - ( \frac{ \sqrt{2} + 1 }{2} ) = ( \frac{ \sqrt{2} - 1 }{2} ) \: \: or \: \: x - ( \frac{ \sqrt{2} + 1 }{2} ) = - ( \frac{ \sqrt{2} - 1 }{2} ). \\ \\ = > x = ( \frac{ \sqrt{2} + 1 }{2} ) + ( \frac{ \sqrt{2} - 1 }{2} ) \: \: \: or \: \: \: x = ( \frac{ \sqrt{2} + 1 }{2} ) - ( \frac{ \sqrt{2} - 1 }{2} ). \\ \\ = > x = \frac{ \sqrt{2} + \cancel1 + \sqrt{2} - \cancel1 }{2} \: \: \: or \: \: \: x = \frac{ \cancel{\sqrt{2} } + 1 - \cancel{\sqrt{2} } + 1}{2} . \\ \\ = > x = \frac{ \cancel2 \sqrt{2} }{ \cancel2} \: \: or \: \: x = \cancel{\frac{2}{2} }. \\ \\ \boxed{ \bf \therefore x = \sqrt{2} \: \: or \: \: x = 1.}$



✔✔ Hence, it is solved ✅✅.




THANKS



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