Can someone put into words the idea of why the solution with the red arrow next to is not included?
Thanks
You must look in the ORIGINAL equation to determine final solutions.
In this case, since you are subtracting 3, the rest of the left-hand side had better be positive or you cannot get zero (0). Further, since the absolute value is positive, this constrains the remaining piece, x, to be positive.
I note two things:
1) 3s are wild. You could simplfy your life by a FIRST step...
x|3x+2| - 1 = 0
2) You did it twice. What was the point of moving the 3 to the right-hand side, just to move it back to the left so you could factor the whole expression?
If I were to do this problem, I might proceed like this:
A) x|9x+6| - 3 = 0
That three (3) is everywhere.
B) x|3x+2| - 1 = 0
x > 0 (by the argument up front.)
If 3x+2 > 0 ==> x > -3/2, and since we already know x > 0, we're just about done.
C) x(3x+2) - 1 = 0
Distributive Property
D) 3x^2 + 2x - 1 = 0
It has one positive solution. (Why do I know that? It's a visual inspection.)
2^2 + 4(3)(1) = 4 + 12 = 16 <== Whoops. I should have been able to factor it.
E) (3x - 1)(x + 1) = 0
F) x = 1/3 or x = -1 (and x = -1 is already thrown out.)
No.
If you like a more formal approach use the definition of the absolute value:
$\displaystyle |3x+2|=\left\{\begin{array}{rcl}3x+2&~if~&3x+2 \geq0~\implies~\boxed{x\geq -\frac23} \\ \\ -(3x+2) &~if~& 3x+2 < 0~\implies~\boxed{x< -\frac23}\end{array}\right.$
Therefore you have to solve for x:
$\displaystyle \left|\begin{array}{rcl}3x^2+2x-1=0 &~\wedge~& x\geq-\frac23 \\ \\ -3x^2-2x-1=0 &~\wedge~& x< -\frac23 \end{array}\right.$
To answer your specific question, -1/3 is not a solution to the equation because it does NOT satisfy the equation!
If x= -3/3= -1, then x|9x+ 6|+ 3= (-1)(9(-1)+ 6|+ 3= (-1)(-9+ 6)+ 3= (-1)(-3)+ 3= 3+ 3= 6, not 0.
Another way of looking at it is that $\displaystyle |x|= \sqrt{x^2}$ so that removing the absolute value is like squaring both sides of the equation- and that can introduce new values of x that satisfy the new equation but not the original one.