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Math Challenge II algebra 7.28

 
 
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Re: Math Challenge II algebra 7.28
LensmireJohn - 2021年10月4日 Monday 12:40
 

Note that sometimes copying and pasting gets a little weird in the forums. It's best to try to right click and "paste as plain text" or first copy the text into a program like "notepad" without any formatting.

As you mentioned, once you know that the maximum root is $x =\sqrt[3]{2} / 2 = 2^{-2/3}$ you can plug it back into the original equation to get an expression in terms of $a$: $$(2^{-2/3})^2 + a^2\cdot 2^{-2/3} + a = 2^{-2/3} a^2 + 1 a + 2^{-4/3} = 0.$$ Note that this is a quadratic in a, which can always be solved using the quadratic formula.

Hint: This isn't as bad as it seems, because the discriminant is $1^2 - 4\cdot 2^{-2/3}\cdot 2^{-4/3} = 0$.