Solve polynomial equation with real parameter

Question

Solve the equation $x^4-(2m+1)x^3+(m-1)x^2+(2m^2+1)x+m=0,$ where $m$ is a real parameter.

My work: So far I've been able to factor the polynomial to $(-x^2+x+m)(-x^2+2mx+1)=0$. Then after using the quadratic formula with each of the factors I'm here: $x=\frac{-1 \pm \sqrt{1+4m}}{-2}$ and $x=\frac{-2m \pm \sqrt{4m^2+4}}{-2}$

Okay, what more do you want? It looks like you have solved the problem! — user247327, Sep 04 '18 at 22:01
Dave, do you need to determine number of real solutions? The second fraction gives always two different real solutions. In the first one, the number of solutions depends on the sign of discriminant $1+4m.$ — user376343, Sep 04 '18 at 22:12
I wasn't sure if you could establish values for m. Thank ya'll — Dave Ezarik, Sep 04 '18 at 22:27

score 0 · Accepted Answer · answered Sep 04 '18 at 22:09

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Take the negatives out. It will look tidier. So: $$(x^2-x-m)(x^2-2mx-1)=0$$ Then: $$x=\frac{1\pm\sqrt{1-4m}}{2}=\frac 12\pm\frac{\sqrt{1-4m}}{2}$$ And: $$x=\frac{2m\pm\sqrt{4m^2+4}}{2}=m\pm\sqrt{m^2+1}$$ You can't do anything more here.

answered Sep 04 '18 at 22:09

Rhys Hughes

12,842

score 0 · Answer 2 · answered Sep 04 '18 at 22:12

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It seems that you are not sure if they are all roots. The fundamental theorem of algebra implies that a polynomial of positive degree $n$ has exactly $n$ roots. In your case, there should be 4 roots and you have found them.

answered Sep 04 '18 at 22:12

Przemysław Scherwentke

13,668
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Solve polynomial equation with real parameter

2 Answers2