Answer :
[tex]2x=\sqrt{3x-1}\implies (2x)^2=(\sqrt{3x-1})^2\implies 4x^2=3x-1\implies 4x^2-3x+1=0[/tex]
You can use the quadratic formula to solve for [tex]x[/tex], but first check the discriminant:
[tex]\Delta(ax^2+bx+c)=b^2-4ac\implies \Delta(4x^2-3x+1)=-27[/tex]
Since the discriminant is negative, this quadratic equation has two complex solutions, so there is no real solution to the original equation.
You can use the quadratic formula to solve for [tex]x[/tex], but first check the discriminant:
[tex]\Delta(ax^2+bx+c)=b^2-4ac\implies \Delta(4x^2-3x+1)=-27[/tex]
Since the discriminant is negative, this quadratic equation has two complex solutions, so there is no real solution to the original equation.