1 Answer

Alex Ross · Answer 1 · 2018-12-15T20:21:54+0000

$2x=√(3x-1)\implies (2x)^2=(√(3x-1))^2\implies 4x^2=3x-1\implies 4x^2-3x+1=0$

You can use the quadratic formula to solve for

, but first check the discriminant:

$\Delta(ax^2+bx+c)=b^2-4ac\implies \Delta(4x^2-3x+1)=-27$

Since the discriminant is negative, this quadratic equation has two complex solutions, so there is no real solution to the original equation.

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