Question

pls help me I will mark u as brainliest

Answer 1

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

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

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

It is in the form of ax^2 + bx + c = 0. The solutions are:

(1)

$x = \frac{-b + \sqrt{b^2 - 4ac} }{2a}$

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

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

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

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

$x = \frac{2 \sqrt{2} }{2}$

$x = \sqrt{2}$



(2)

$x = \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

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

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

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

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

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

$x = \frac{2}{2}$

x = 1.



$Therefore : x = \sqrt{2} , x = 1$



Hope this helps!

Answer 2

Hi,

Please see the attached file!


Thanks