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Question
`x+1/x=3,x≠0`
Solution
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.
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