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The Equation $x^4+2x^3+mx^2+2x+1=0$ has $4$ different real roots iff:
a) $m<3$;
b) $m<2$;
c) $m<-6$;
d) $1<m<3$;
e) $-6<m<2$

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What do you think? How do you plan to approach it? –  vonbrand Feb 7 at 17:25
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Sure it's not $mx^2$ instead of $mx$? –  Hagen von Eitzen Feb 7 at 17:48
    
Yes, sorry. I will correct it! –  veljovic Feb 7 at 17:50

1 Answer 1

up vote 3 down vote accepted

Notice that $x=0$ is not going to be a solution. Then notice the coefficients are symmetric. This means that if we divide by $x^{4/2}$ we get $$(x^2+1/x^2)+2(x+1/x)+m=0$$

which can be written as

$$(x+1/x)^2+(x+1/x)+m-2=0.$$

Putting $y:=x+1/x$ we get $$y^2+y+m-2=0$$

which has two different real solutions for $1-4(m-2)>0$, i.e. $9>4m$. For these we have $x+1/x=y=\frac{-1\pm\sqrt{1-4(m-2)}}{2}$.

These gives us two polynomials of degree two:

$$x^2-\frac{-1\pm\sqrt{1-4(m-2)}}{2}x+1.$$

To get two different real solutions out of these we need again positive discriminant $\left(\frac{-1\pm\sqrt{1-4(m-2)}}{2}\right)^2-4>0$.

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