4(x-3)^2 +1=√((x-1)/4) +3
4(x-3)^2 - 2 = √((x-1)/4) = 1/2 * √(x-1)
8(x-3)^2 - 4 = √(x-1)
8 x^2 - 48 x + 68 = √(x-1) = t
8 (t^2 + 1)^2 - 48 (t^2 + 1) + 68 = t (because x = t^2 + 1)
8 t^4 - 32 t^2 - t + 28 = 0
I am stuck here.
i need a help solving this algebraic equation
4(x-3)^2 +1=√((x-1)/4) +3
plz i need some quick answers
i would really appreciate whoever solves this question
thanks