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 07-02-2012, 11:02 AM #31 newbie   Join Date: Aug 2006 Posts: 31 Re: Need help with a basic algebra problem??? It seems as though the question was carelessly formulated. Given any x, letting y be equal to (Ax^2 - 8) obviously yields a "solution". However, what the teacher presumably meant to ask was "For what values of A does the function y = Ax^2 - 8 lack roots?", which means "For what values of A does the equation Ax^2 - 8 = 0 lack solutions?". The latter question is silly if we use complex analysis, so let's look only for real-valued solutions: Obviously if A = 0, there is no solution. Otherwise our equation is equivalent to x^2 = 8/A. Squares of real numbers are always non-negative. Thus if 8/A < 0, i.e. iff A < 0, then there is no solution. And if 8/A > 0, i.e. iff A > 0, then there are the two solutions x = \sqrt{8/A} and x = - \sqrt{8/A}.
07-02-2012, 02:07 PM   #32
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Join Date: Mar 2011
Posts: 4,448
Re: Need help with a basic algebra problem???

Quote:
 Originally Posted by Tlick It seems as though the question was carelessly formulated. Given any x, letting y be equal to (Ax^2 - 8) obviously yields a "solution". However, what the teacher presumably meant to ask was "For what values of A does the function y = Ax^2 - 8 lack roots?", which means "For what values of A does the equation Ax^2 - 8 = 0 lack solutions?". The latter question is silly if we use complex analysis, so let's look only for real-valued solutions: Obviously if A = 0, there is no solution. Otherwise our equation is equivalent to x^2 = 8/A. Squares of real numbers are always non-negative. Thus if 8/A < 0, i.e. iff A < 0, then there is no solution. And if 8/A > 0, i.e. iff A > 0, then there are the two solutions x = \sqrt{8/A} and x = - \sqrt{8/A}.

Yeah, this is a good answer. 0 is the only real number that will cause issues with a for this reason above and you can think about the fact that if a = 0, there are infinite numbers that can be used as an x value and you are only supposed to have two roots.

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