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Square in an ellipse, circle not so much... Unless I misunderstand the question, no. See the sketch on the right for a counterexample of the 2d-case. --Stephan Schulz (talk) 12:32, 11 March 2017 (UTC)[reply]
@31.154.81.69: Your equation makes no sense, since the term bx is not a scalar when b is a mere scalar. If you mean for b to be instead a covector, then this is not true in general unless A is a positive-definite matrix.--Jasper Deng(talk)05:53, 12 March 2017 (UTC)[reply]
Actually, it is not even enough that A be positive-definite. The whole quadratic form has to be positive definite, or the bx term can ruin things (consider the value on -kei for a large scalar k).--Jasper Deng(talk)19:34, 12 March 2017 (UTC)[reply]
Yes, the statement is not true. It appears to be a homework problem, so I don't want to answer in detail, but 2 dimensional counterexamples are easy to come by. A hint: if is a solution, then is in the span of the solutions (the counterexample I came up with uses a large b rather than a large k).--Wikimedes (talk) 03:36, 13 March 2017 (UTC)[reply]