Let $$n$$ be a rational number such that the equation
$$x^2-(\sqrt{3}+1)x+\sqrt{3}n-6 = 0$$
has an integer root. Find the value of $$n$$.

Question

Let $$n$$ be a rational number such that the equation 
$$x^2-(\sqrt{3}+1)x+\sqrt{3}n-6 = 0$$
has an integer root. Find the value of $$n$$.

Uniteasy Admin · Accepted Answer

$$x^2-(\sqrt{3}+1)x+\sqrt{3}n-6 = 0$$c
Reorganize the terms of the equation to move irrational terms to the RHS
$$x^2-x-6 = \sqrt{3}x-\sqrt{3}n$$c
Since $$\sqrt{3} $$ is an irrational number, in order to make the LHS rational, RHS must meet the condition
$$x -n = 0$$nc
Then the following equation is obtained
$$x^2-x-6 = 0$$nc
Solving the quadratic equation yields two roots
$$x=3 	ext{ or } x=-2$$c
According to the condition (1), the values of $$n$$ is the same as $$x$$
$$n=3 	ext{ or } nx=-2$$c

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