How to solve this system with equation and inequality? { <mtable columnalign="right

migongoniwt

migongoniwt

Answered question

2022-06-21

How to solve this system with equation and inequality?
{ 2 x + 2 = 49 4 x 2 + 4 , 2 x + 2 4 x 2 ( 14 2 x + 2 ) 2 x

Answer & Explanation

Elianna Douglas

Elianna Douglas

Beginner2022-06-22Added 23 answers

Well, as suspected, your edit brings a new piece of information (the inequation) that allows to easily solve that system using only high-school mathematics (the role of the inequality being to rule out the "ugly" non-zero solutions) - with the exception of a single step in the proof that seems to require very high-level mathematics.
For simplicity, let t = 2 x . The equation implies that x 2 = 16 49 ( t 1 ). Plugging this into the inequality and performing all the necessary simplifications brings it to
( t 1 ) ( 49 56 t + 16 t 2 ) 0 .
Inside the second pair of brackets one recognizes ( 7 4 t ) 2 , so there are only two possibilities:
- either t 1 0 and ( 7 4 t ) 2 0
- or t 1 0 and ( 7 4 t ) 2 0
Since ( 7 4 t ) 2 0, the first possibility implies that ( 7 4 t ) 2 = 0 so that t = 7 4 , which also satisfies t 1 0. Returning to x, this has the consequence that 2 x = 7 4 and x 2 = 12 49 , or equally well x = log 2 7 2 and x = ± 2 3 7 . Since x = log 2 7 2 log 2 4 2 = 0, it follows that x cannot take that negative value. Can it take the positive one? Assuming it could, this would mean that log 2 7 = 2 + 2 3 7 , which after exponentiation would be equivalent to 7 4 = 4 3 7 . This, in turn, implies that 4 3 7 Q, which after raising to the 7th power implies that 4 3 Q. This is not true, but I do not know how to prove this at a high-school level. Maybe you are supposed to use a scientific calculator to solve your problem? Anyway, that 4 3 is irrational is a consequence of the Gelfond-Schneider theorem, or of the Lindemann-Weierstrass theorem, both of them, though, being vastly out of the reach of high-school pupils. In any case, the conclusion so far is that t = 7 4 does not lead to solutions of the given system.
The second possibility reduces to t 1 0, because the square is always 0, but since x 2 = 16 49 ( t 1 ), this would imply that x 2 = 0 ( 0 being the only number both 0 and 0), which has the only solution x = 0.
The analysis could have been performed equally well by replacing t with 49 16 x 2 + 1 and performing all the compuations with x, instead of with t, but the reasoning would have reached the same difficult point about log 2 7 that has been reached above.

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