- #26
underlined part is unclear. I assumed ## b^2=20a## holds and parametrised the expression ##5a+b## with respect to ##b##. The graph of the resulting function ##B## is just a square parabola so there's no secret what its extremum point is.got the function B = b2/4 + b by taking the minimum value of 5a
5a ≥ b^{2}/4 right? So the minimum value of 5a is b^{2}/4. Can't neglect that possibility.underlined part is unclear. I assumed ## b^2=20a## holds and parametrised the expression ##5a+b## with respect to ##b##. The graph of the resulting function ##B## is just a square parabola so there's no secret what its extremum point is.
That is just an accident. The question itself did not specify that ##a## or ##b## need be integers. Likewise, it did not specify that ##5a+b## need be an integer, but possibly with non-integer ##a,b##. Of course, I am assuming you wrote out a complete statement of the question that was given to you.The answer they've given is an integer.
Yes, I did. But B when differentiated gives -1 as the extremum has some connection for sure.That is just an accident. The question itself did not specify that ##a## or ##b## need be integers. Likewise, it did not specify that ##5a+b## need be an integer, but possibly with non-integer ##a,b##. Of course, I am assuming you wrote out a complete statement of the question that was given to you.
b=-2You have the condition ##|b| < 2\sqrt{5a} ## to consider. Given the initial post, ##a \geq 0?## may be arbitrary.
For which value of ##b## is the vertex of the resulting parabola lowest possible?
I'm not sure I'm getting your point.That is just an accident. The question itself did not specify that ##a## or ##b## need be integers. Likewise, it did not specify that ##5a+b## need be an integer, but possibly with non-integer ##a,b##. Of course, I am assuming you wrote out a complete statement of the question that was given to you.
I get it now. A little imagination did the trickTo elaborate more, B ≥ b^{2}/4 + b is a family of parabolas for all values of b and the 'greater than part' represents the locus of the points outside the interval (α,β) if they're the roots of B
how do you order arbitrary complex numbers? The expression ##(x,y) ## as an interval makes sense in this context if they are comparable.I get it now. A little imagination did the trick
My computer isn't recognising these as symbols, so they're just visible as plain text. Do you mean to say that 5a + b is an integer does not imply a,b are intergers?What he means is in general ##5a+b\in\mathbb Z\implies a,b\in\mathbb Z ## is false.
complex numbers or not, for simpliclity, I imagined thishow do you order arbitrary complex numbers? The expression ##(x,y) ## as an interval makes sense in this context if they are comparable.
Is it 35? Just to know whether I've got the right answer before I make a fool of myself telling everyone it's easy!The answer they've given is an integer.
Please respond using words. I cannot figure out what you are attempting to convey here, and likewise, I cannot figure out which post you are replying to.
Nothing, just reacting to epenguin's post (#44). Humor--plain and simplePlease respond using words. I cannot figure out what you are attempting to convey here, and likewise, I cannot figure out which post you are replying to.