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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 $\displaystyle \sqrt{x^4-6x^2+9}-3a\sqrt{x^2-3}+2a^2=0$ 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} Thanks from topsquark March 9th, 2019, 09:12 PM   #3
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Quote:
 Originally Posted by v8archie \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}
Hello,

Are you kidding?What is unclear?The problem I posted is from another forum.

All the best,

Integrator March 9th, 2019, 09:45 PM #4 Senior Member   Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 $\sqrt{x^4-6x^2+9}-3a\sqrt{x^2-3}+2a^2 = 0$ $\sqrt{(x^2-3)^2}-3a\sqrt{x^2-3}+2a^2 = 0$ let $u = \sqrt{x^2-3}$ $u^2 - 3au + 2a^2 = 0$ 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. $x = \pm \sqrt{3+a^3},~\pm \sqrt{3+8a^3}$ Thanks from topsquark March 10th, 2019, 09:36 AM   #5
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Quote:
 Originally Posted by Integrator Are you kidding?What is unclear?The problem I posted is from another forum.
So what? The suggestion I made came from my own head. If you want the solution, why not look on the other forum? March 10th, 2019, 09:33 PM   #6
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Quote:
 Originally Posted by romsek $\sqrt{x^4-6x^2+9}-3a\sqrt{x^2-3}+2a^2 = 0$ $\sqrt{(x^2-3)^2}-3a\sqrt{x^2-3}+2a^2 = 0$ let $u = \sqrt{x^2-3}$ $u^2 - 3au + 2a^2 = 0$ 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. $x = \pm \sqrt{3+a^3},~\pm \sqrt{3+8a^3}$
Hello,

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 Tags équation, equation, une Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post max233 Algebra 2 March 31st, 2017 09:40 PM DarkLink94 Algebra 5 July 4th, 2015 02:32 PM DarkX132 Algebra 3 September 26th, 2014 10:15 PM Sonprelis Calculus 6 August 6th, 2014 10:07 AM PhizKid Differential Equations 0 February 24th, 2013 10:30 AM

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