Question

`x^2-4x-1=0` 

Answer 1

Given: 

`x^2-4x-1-=0` 

On comparing it with `ax^2+bx+c=0` 

`a=1,b=-4 and c=-1` 

Discriminant D is given by: 

`D=(b^2-4ac)`  

=`(-4)^2-4xx1xx(-1)` 

=`16+4` 

=`20` 

=`20>0`  

Hence, the roots of the equation are real.
Roots α  and β are given by: 

α=`(-bsqrt(D))/(2a)=(-(-4)+sqrt(20))/(2xx1)=(4+2sqrt(5))/2=(2(2+sqrt(5)))/2=(2+sqrt(5))` 

β=`(-b-sqrt(D))/(2a)=(-(-4)-20)/2=(4-2sqrt5)/2=(2(2-sqrt(5)))/2=(2-sqrt(5))` 

Thus, the roots of the equation are` (2+sqrt(5)) and (2-sqrt(5))`