Question

.......Solve the equation:

Answer 1

Notice that the factored form of the part under the radical is:

(x-3)(x-3) which is equal to:

(x-3)^2

√((x-3)^2)+3|(x-3)(2x+1)|=0

√((x-3)^2)=-3|(x-3)(2x+1)|  now squaring both sides

(x-3)^2=9(x-3)^2*(2x+1)^2  dividing both sides by (x-3)^2

1=9(2x+1)^2

1=9(4x^2+4x+1)

1=36x^2+36x+9

36x^2+36x+8=0

36x^2+36x=-8

x^2+x=-2/9

x^2+x+1/4=1/4-2/9

(x+1/2)^2=(9-8)/36

(x+1/2)^2=1/36  taking the square root of both sides

x+1/2=±1/6

x=-1/2±1/6

x=(-3±1)/6

x= -2/3 and -1/3 and of course 3 because of the initial condition...

x=-1/3, -2/3, 3