Answer:
C. x =3
Explanation:
Extraneous solution is that root of a transformed equation that doesn't satisfy the equation in it's original form because it was excluded from the domain of the original equation.
Let's solve the equation first
![√(45-3x) = x-9\\Taking\ square\ on\ both\ sides\\{(√(45-3x))}^2 = {(x-9)}^2\\45-3x = x^2-18x+81\\0 = x^2-18x+81-45+3x\\x^2-15x+36 = 0\\x^2-12x-3x+36 = 0\\x(x-12)-3(x-12) = 0\\(x-3)(x-12)\\x-3 = 0\\=> x =3\\x-12 = 0\\x = 12\\We\ will\ check\ the\ solutions\ one\ by\ one\\So,\\for\ x=3\\√(45-3(3)) = 3-9\\√(45-9) = -6\\√(36)= -6\\6\\eq -6\\For x=12\\√(45-3(12)) = 12-9\\√(45-36) = 6\\√(36)= 6\\6=6](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ktdq4av4pxpr6jdtb1o6an6r5td6yvqkrx.png)
Hence, we can conclude that x=3 is an extraneous solution of the equation ..