Am I crazy, or are the roots of x^2+2x +1 (x+1)(x+1), NOT (x-1)(x-1)
This would make A correct not C. That’s how I got to my answer.
(x + 1)(x + 1) has root of multiplicity 2 at x = -1, so i’m not sure what you mean. This doesn’t contradict what’s written there, does it?
Not sure I follow- what does root of multiplicity mean?
Is this different than factoring out x^2+2x+1 ?
(x + 1)^2 has roots at x = -1 right? Isn’t that what the solution says too?
By roots do you mean (x+1)^2 = 0 when x = -1?
Lol instead of responding, i ended up editing my previous comment by accident, but anyway this was my response:
Yes, I do mean that
The solutions says that if \alpha & \beta represent your two roots, then \alpha = \beta = -1. This is not disjoint from what you said because x^2 + 2x + 1 = (x + 1)^2 = 0 \implies x = -1, where x happens to be your root(s). To reiterate yeah, you’re just misinterpreting the solution and it is in line with what you said.
From here on out, you don’t have to read unless you’re curious about the multiplicity thing I mentioned prior, so here goes:
Technically there’s only one root but with multiplicity 2 (two copies). In other words, if we consider our polynomial (but with a slight tweak) as:
(x + 1)(x + 1 - \epsilon)
where \epsilon is a “super small” quantity but not 0. It should be clear that this quadratic has two roots, namely: -1 and -1 + \epsilon. Now you can imagine this “super small quantity” (\epsilon) approaching 0, and this should thus intuitively signal how the two roots coincide, thus leaving you with only one root but double copies. This “double” is what we can refer to as a multiplicity of 2 or a double root.
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Thank you! That is extremely helpful. Seems to be just a Math vocab issue, since I wasn’t sure what the roots of a problem were.