Question

`x^2-(sqrt2+1)x+sqrt2=0` 

Answer 1

`x^2-(sqrt2+1)x+sqrt2=0` 

⇒`x^2-(sqrt2+1)x=-sqrt2`  

⇒`x^2-2xx x xx((sqrt2+1)/2)+((sqrt2+1)/2)^2=sqrt2+((sqrt2+1)/2)^2` 

[Adding `((sqrt2+1)/2)]^2` on the sides] 

⇒`[x-((sqrt2+1)/2)]^2=(-4sqrt2+2+1+2sqrt2)/4=(2-2sqrt2+1)/4=((sqrt2-1)/2)^2` 

⇒`x-((sqrt2+1)/2)=+-((sqrt2-1)/2)`        (Taking square root on both sides) 

⇒`x-((sqrt2+1)/2)=((sqrt2-1)/2) or  x-((sqrt2-1)/2)=-((sqrt2-1)/2)` 

⇒`x=(sqrt2+1)/2+(sqrt2-1)/2 or  x=(sqrt2+1)/2-(sqrt2-1)/2` 

⇒`x=(2sqrt2)/2=sqrt2 or  x=2/2=1` 

Hence, `sqrt2 and 1` are the roots of the given equation.