Question

How to solve for X for 2X= 3X-1 when 3X-1 is under a square root

Answer 1

`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
`x`, 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.