Solve x^2=3(x-1)^2

To start with, we want to put this into quadratic form, where we have ax^2+bx+c=0. We notice that there are terms on both sides of the equation. So we can expand the right hand side and then bring all the terms to one side. To make it easier to visualise and expand , we can write (x-1)^2 as (x-1)(x-1). If we take RHS as 3(x-1)^2 we have,RHS: 3(x-1)(x-1)=3(x^2-x-x+1)= 3(x^2-2x+1)=3x^2-6x+3
Now we subtract the LHS x^2 from both sides. So 2x^2-6x+3=0. To factorise this we want to find two terms that multiply to make 3 and add together to make six. There is no combination of integers that do this so we use the quadratic equation formula. This gives x=(-(-6)±sqrroot((-6)^2-4(2)(3))/2(2) =(6 ±sqrroot(12))/4. So x=2.37 and x=0.63 (2.d.p)