Question

solve for x: x square minus (root 2 + 1 ) X + root 2 equals to zero

Answer 1

hope this will help you

Answer 2

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

We need to solve for x.

On splitting the middle term, we get

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

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

Taking common from each parentheses, we get

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

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

Using zero product property, we get

$x-\sqrt{2}=0\text{ or }x-1=0$

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

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