Question
Solve the radical equation
\sqrt{2x-1}+\sqrt{4x-2}+\sqrt{6x-3} = 6x-3
Solve the radical equation
\sqrt{2x-1}+\sqrt{4x-2}+\sqrt{6x-3} = 6x-3
\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
or
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