Online Course Discussion Forum

Math Challenge II algebra 7.28

 
 
Picture of Jialin Wang
Math Challenge II algebra 7.28
by Jialin Wang - Friday, October 1, 2021, 6:29 PM
 

i got cube root of 2/2 and they asked my to find a an I have no idea how to perform the calculations. How do I persorm the calculations?

 
Picture of Jialin Wang
Re: Math Challenge II algebra 7.28
by Jialin Wang - Friday, October 1, 2021, 7:22 PM
 

We know that the maximum root is 232232. If this is a root we have

(232)2+a2(232)+a=0.(232)2+a2(232)+a=0.
Solving for aa gives a=123=2322a=−123=−2322.


It's the answer but i dont know how to solve for a in this case

Picture of John Lensmire
Re: Math Challenge II algebra 7.28
by John Lensmire - Monday, October 4, 2021, 12:40 PM
 

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 $1^2 - 4\cdot 2^{-2/3}\cdot 2^{-4/3} = 0$.