# Q. Consider the equation `x^3-nx+1=0;n in N;n>=3.` Then A)Equation has atleast one rational root. B)Equation has exactly one rational root. C)Equation has atleast one root belonging to (0,1)....

Q. Consider the equation `x^3-nx+1=0;n in N;n>=3.` Then

A)Equation has atleast one rational root.

B)Equation has exactly one rational root.

C)Equation has atleast one root belonging to (0,1).

D)Equation has no rational root.

[Ans:A but how??]

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Given the equation `x^3-nx+1=0,n in NN` determine the number of rational roots for `n>=3` :

By the rational root theorem, any possible rational root is of the form `r=p/q` where r is the root, p is a factor of the constant term, and q is a factor of the leading coefficient.

For `x^3-nx+1` the leading coefficient is 1, and the constant term is 1. The only possible rational roots are `r=+-1` .

We can check the roots using synthetic division:

1 | 1 0 -n 1

1 1 -n+1

-----------------

1 1 -n+1 -n+2

`-n+2!=0` if `n>=3` so 1 is not a root.

-1 | 1 0 -n 1

-1 1 n-1

----------------

1 -1 -n+1 n

`n!=0` if `n>=3` so -1 is not a root.

**Using the rational root theorem, we see that there are no rational roots for `x^3-nx+1=0` if `n>=3` , so the answer is D.**

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` `A and B are wrong since there cannot be any rational roots.

**C is also correct**: for `n>=3` there is always a real root in (0,1).

For `n>=3` the discriminant is positive so there are three distinct real roots -- we already know they cannot be rational. At x=0 the function has the value 1. At x=1 the function has the value -n+2 which is negative for `n>=3` ; since the function is continuous and the graph goes from positive values to negative values, it must be zero in the interval.