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March 9th, 2019, 01:14 AM  #1 
Member Joined: Aug 2018 From: România Posts: 85 Thanks: 6  Re: An equation
Hello all, Determine the set of all values of the parameter $\displaystyle a\geq 0$ knowing that the equation $\displaystyle \sqrt[3]{x^46x^2+9}3a\sqrt[3]{x^23}+2a^2=0$ has all the real roots. All the best, Integrator 
March 9th, 2019, 02:56 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra 
\begin{align} a+b&=c \\ (a+b)^3 &= c^3 \\ a^3 + a^2b + ab^2 + b^3 &= c^3 \\ a^3 + (a+b)ab + b^3 &= c^3 \\ a^3 +abc + b^3 &= c^3 \end{align} 
March 9th, 2019, 09:12 PM  #3 
Member Joined: Aug 2018 From: România Posts: 85 Thanks: 6  
March 9th, 2019, 09:45 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 
$\sqrt[3]{x^46x^2+9}3a\sqrt[3]{x^23}+2a^2 = 0$ $\sqrt[3]{(x^23)^2}3a\sqrt[3]{x^23}+2a^2 = 0$ let $u = \sqrt[3]{x^23}$ $u^2  3au + 2a^2 = 0$ The discriminant $D = (3a)^2  4(2a^2) = a^2$ $\forall a \geq 0,~D\geq 0$ Thus the equation will always have real roots. $x = \pm \sqrt{3+a^3},~\pm \sqrt{3+8a^3}$ 
March 10th, 2019, 09:36 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra  
March 10th, 2019, 09:33 PM  #6  
Member Joined: Aug 2018 From: România Posts: 85 Thanks: 6  Quote:
Some say that the equation has all the real roots only if $\displaystyle a=0$ , because the equation has $\displaystyle 12$ roots of which $\displaystyle 4$ are real and $\displaystyle 8$ complex roots. What are the complex roots? All the best, Integrator  

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