Question

`x+1/x=3,x≠0` 

Answer 1

The given equation is 

`x+1/x=3,x≠0` 

⇒` (x^2+1)/x=3 ` 

⇒ `x^2+1=3x` 

⇒`x^2-3x+1=0` 

This equation is of the form `ax^2+bx+c=0,` where, `a=1, b=-3 and c=1` 

∴ Discriminant,` D=b^2-4ac=(-3)^2-4xx1xx1=9-4=5>0` 

 

So, the given equation has real roots.
Now, `sqrtD=sqrt5` 

∴`a=(-b+sqrtD)/(2a)=(-(-3)+sqrt5)/(2xx1)=(3+sqrt5)/2` 

β=(-b-sqrtD)/(2a)=(-(-3)-sqrt5)/(2xx1)=(3-sqrt5)/2`  

Hence, `(3+sqrt5)/2` and `(3-sqrt5)/2` are the roots of the given equation.