Plus and Minus and Completeing the Square/Cube

[dtmowns]

I think I understand the use of 'plus and minus' before the square root when completing the square and using the quadratic formula: am I correct in saying it is simply a way to account for the different combinations of x-intercepts for a parabola (i.e. both a "left" and a "right" intercept, one intercept at the origin, or no intercepts)? When you have a quadratic set equal to zero, the solutions to that quadratic are the x values when y is equal to zero, correct?

My main question is about using "plus and minus" with third degree polynomials. Here is an example:

*Given that x^3 + 3x^2 + 3x + 1 = (x+1)^3, find all real x such that x^3 + 3x^2 + 3x = 1.

Now, I understand the operation of adding 1 to both sides, and then replacing the left side with (x+1)^3, and then taking the cube root of both sides to get x + 1 = the cube root of 2, but my inability to decide whether a plus and minus should be placed in front of the right side at this point makes me think I need to broaden my understanding. In this case, a third degree polynomial is set equal to 2. Does this mean that the solutions to that third degree polynomial are the x values when y is equal to 2? If so, wouldn't you need a 'plus and minus'? I find that not enough emphasis is placed on tying these equations to their graphs in instructional books.

 

[MarkFL]

We use the ± sign when taking an even root, to account for the fact that:

\(\displaystyle (\pm x)^{2n}=x^{2n}\)

However, when we are taking an odd root, we find:

\(\displaystyle (\pm x)^{2n+1}=\pm x^{2n+1}\)

Hence we find the sign must be preserved. For example,

\(\displaystyle x^2=4\implies x=\pm4\) whereas \(\displaystyle x^3=-8\implies x=-2\) (for real values of \(\displaystyle x\)).

So, in the case of the cubic equation you cite, you would not attach the ± sign, as there is only 1 real root, the other two roots are complex conjugates.