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.

Collected in the board: Quadratic function

Steven Zheng posted 1 week ago

Answer

x^2-(\sqrt{3}+1)x+\sqrt{3}n-6 = 0

Reorganize the terms of the equation to move irrational terms to the RHS

x^2-x-6 = \sqrt{3}x-\sqrt{3}n

Since \sqrt{3} is an irrational number, in order to make the LHS rational, RHS must meet the condition

x -n = 0
(1)

Then the following equation is obtained

x^2-x-6 = 0
(2)

Solving the quadratic equation yields two roots

x=3 \text{ or } x=-2

According to the condition (1), the values of n is the same as x

n=3 \text{ or } nx=-2

Steven Zheng posted 1 week ago

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