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So, I got some tasks from my lecturer about quartic and even quintic equation

Here is the question

I tried using some general solution for quartic from this link

Is there a general formula for solving 4th degree equations (quartic)?

I failed and the equation become more and more complex. Anyone can help me with this?

Zulvan Avivi
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1 Answers1

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Solving $$x^4 + 4 a x^3 + 6 x^2 + 4 a x + 1 = 0$$ Divide all terms by $x^2$ $$x^2+4ax+6+\frac{4a}{x}+\frac{1}{x^2}=0$$ Osserve that we can write the previous equation as $$\left(x^2+\frac{1}{x^2}+2\right)+4a\left(x+\frac{1}{x}\right)+4=0$$ That is $$\left(x+\frac{1}{x}\right)^2+4a\left(x+\frac{1}{x}\right)+4=0$$ Set $w=x+\frac{1}{x}$. The equation becomes $$w^2+4aw+4=0$$ Solutions are $$w_1=-2a-2 \sqrt{a^2-1};\;w_2=-2a+2 \sqrt{a^2-1}$$ Bringing beack the $x$ we have the equations $$x+\frac{1}{x}=-2a-2 \sqrt{a^2-1};\;x+\frac{1}{x}=-2a+2 \sqrt{a^2-1}$$ Reordering etc $$x^2 + \left(2a+2 \sqrt{a^2-1}\right)x+1=0;\;x^2 + \left(2a-2 \sqrt{a^2-1}\right)x+1=0$$ Whose solutions are $$x_1=-a-\sqrt{a^2-1}-\sqrt{2 a^2-2+2 a\sqrt{a^2-1} };\;\\x_2=-a-\sqrt{a^2-1}+\sqrt{2 a^2-2+2 a\sqrt{a^2-1}};\;\\x_3=-a+\sqrt{a^2-1}-\sqrt{2 a^2-2-2 a\sqrt{a^2-1} };\;\\x_4=-a+\sqrt{a^2-1}+\sqrt{2 a^2-2-2 a\sqrt{a^2-1} }$$

Raffaele
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  • It somehow doesn't match with the answer given by my lecturer, but I think your work is so clear and I can't find anything wrong with it. So may it possible for both answer be right ? – Zulvan Avivi Oct 07 '20 at 02:10
  • @ZulvanAvivi They are. I checked and they give the same results for any $a$. By the way, the solutions are pairwise reciprocal,i.e., their product is $1$. – Raffaele Oct 07 '20 at 07:14
  • I see, thank you so much for your clear explanation. – Zulvan Avivi Oct 07 '20 at 07:40