Solving for x in a (modified) quadratic \(\displaystyle{\left(\sqrt{{{49}+{20}\sqrt{{6}}}}\right)}^{{\sqrt{{{a}\sqrt{{{a}\sqrt{{{a}\cdots\infty}}}}}}}}+{\left({5}-{2}\sqrt{{{6}}}\right)}^{{{x}^{{2}}+{x}-{3}-\sqrt{{{x}\sqrt{{{x}\sqrt{{{x}}}\cdots\infty}}}}}}={10}\) where



Answered question


Solving for x in a (modified) quadratic
where   where  a=x23,(a0),  Solve for   x

Answer & Explanation

Harry Gibson

Harry Gibson

Beginner2022-04-01Added 13 answers

There is nothing wrong but if x=±2 then a=1<0 so aa is not well-defined. Hence, x=±2 must be "disqualified" as a solution.
Usually it is always good to check all solutions you got with the original equation since we usually only prove the direction that if x satisfies then equation then x is equal to this and this, but we actually want to prove the other direction as well.

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