Question

Solve the radical equation

\sqrt{2x-1}+\sqrt{4x-2}+\sqrt{6x-3} = 6x-3

Collected in the board: Radical equations

Steven Zheng posted 17 hours ago

Answer

\sqrt{2x-1}+\sqrt{4x-2}+\sqrt{6x-3} = 6x-3
(1)

\sqrt{2x-1}+\sqrt{2(2x-1)}+\sqrt{3(2x-1)} = 3(2x-1)

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

therefore

\sqrt{2x-1} = 0
(2)

or

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

From (2), we get one solution

x = \dfrac{1}{2}

From (3), we get

1+\sqrt{2}+\sqrt{3}=3\sqrt{2x-1}

(1+\sqrt{2}+\sqrt{3})^2 =18x-9

x = \dfrac{(1+\sqrt{2}+\sqrt{3})^2+9}{18} \approx 1.46

Substitute the result to (1) to verify, which shows this is another solution to the equation,

In suumary, there are two solutions

x = \dfrac{1}{2} and x = \dfrac{(1+\sqrt{2}+\sqrt{3})^2+9}{18} \approx 1.46

Steven Zheng posted 17 hours ago

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