Question

`x^2-(sqrt3+1)x+sqrt3=0` 

 

Answer 1

The given equation is `x^2-(sqrt3+1)x+sqrt3=0`  

Comparing it with `ax^2+bx+c=0` we get 

`a=1, b=-(sqrt3+1) and c=sqrt3` 

∴ Discriminant, 

`D=b^2-4ac=[-(sqrt3+1)]^2-4xx1xxsqrt3=3+1+2sqrt3-4sqrt3=3-2sqrt3+1=-2sqrt3+1=(sqrt3-1)^2>0` 

So, the given equation has real roots.
Now,  `sqrtD=sqrt(sqrt3-1)^2=sqrt3-1` 

∴ `α =(-b+sqrt(D))/(2a)=(-[-sqrt(3)+1]+[sqrt(3)-1])/(2xx1)=(sqrt(3)+1+sqrt(3)-1)/2=(2sqrt(3))/2=sqrt3` 

β=`(-b-sqrt(D))/(2a)=(-[-sqrt(3)+1]+[sqrt(3)-1])/(2xx1)=(sqrt(3)+1-sqrt(3)-1)/2=2/2=1` 

 Hence, `sqrt3` and `1` are the roots of the given equation.