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=2
LHS
[tex]\sqrt{2-2}-4\\\\=-4[/tex]
RHS
4-6
= -2
LHS=RHS
Hence, x=2 , is solution of the system.
⇒For , x=3
.LHS=
[tex]\sqrt{3-2}-4\\\\=1-4\\\\=-3[/tex]
RHS
3-6=-3
Hence, x=3 , is solution of the system.
Option C
x=3, is true solution.
Option E:
The zeros of 0 = x2 – 5x + 6 are possible solutions to the radical equation.
