Question

• (i) x² – (1 + root2 )x + root 2 = 0​

Answer 1

Answer:

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

so,. X = 1, √2

Answer 2

Answer:

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

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

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

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

$x = \sqrt{2}$

$x = 1$

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