Which statements describe the solutions to √x-2 -4=x-6 ? Check all that apply. There are no true solutions to the radical equation. x = 2 is an extraneous solution. x = 3 is a true solution. There is only 1 true solution to the equation. The zeros of 0 = x2 – 5x + 6 are possible solutions to the radical equation.

Solution [tex]\sqrt{x-2}-4=x-6\\\\\sqrt{x-2}=x-6+4\\\\\sqrt{x-2}=x-2\\\\ \text{Squaring both sides}\\\\x-2=x^2-4x+4\\\\x^2-4x-x+4+2=0\\\\x^2-5 x+6=0\\\\ \text{Splitting the middle term}\\\\x^2-3x-2x+6=0\\\\x(x-3)-2(x-3)=0\\\\(x-2)(x-3)=0\\\\x=2,3[/tex]Put, x=2 and x=3 in original equation⇒For, x=2LHS  [tex]\sqrt{2-2}-4\\\\=-4[/tex]RHS 4-6= -2LHS=RHSHence, x=2 , is solution of the system.⇒For , x=3.LHS= [tex]\sqrt{3-2}-4\\\\=1-4\\\\=-3[/tex]RHS3-6=-3Hence, x=3 , is solution of the system.Option Cx=3, is true solution.Option E:The zeros of 0 = x2 – 5x + 6 are possible solutions to the radical equation.