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Find all values of a for which the equation $$x^4 +(a-1)x^3 +x^2 +(a-1)x+1=0 $$ possesses at least two distinct negative roots.

I am able to prove that all roots would be negative .How to proceed after this.

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It is not true that all roots must be negative. For example, if $a=1$ there are roots near $0.531$ and $1.88$ and no negative real roots. Here is the Alpha output. – Ross Millikan Jul 15 '13 at 17:18

This is a symmetric (the coefficients) polynomial. We may want to take advantage of that. We can divide by $x^2$ and get $$(x+\frac{1}{x})^2+(a-1)(x+\frac{1}{x})-1=0$$

From there we can solve for $x+\frac{1}{x}$ and get $$x+\frac{1}{x}=\frac{-(a-1)\pm\sqrt{(a-1)^2+4}}{2}.$$

This is a quadratic equation. Solve for $x$ and get the roots.

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It's not necessary to get the roots to find out the values of $a$ for which the equation has two distinct negative roots. – RicardoCruz Jul 15 '13 at 21:40
I know. But it is the OP who might need to know about how to do it. Tell him/her how. – Mlazhinka Shung Gronzalez LeWy Jul 15 '13 at 21:51
@mathslover do you know for which $x<0$, $x+\frac{1}{x}$ assumes a maximum value, and how much is that value? – RicardoCruz Jul 15 '13 at 22:01
max value is -2. @RicardoCruz how to do otherwise – maths lover Jul 16 '13 at 1:39
Well, @mathslover , if you impose $\frac{-(a-1)-\sqrt{(a-1)^2+4}}{2}<-2$ you will get your answer: $a>\frac{5}{2}$ – RicardoCruz Jul 16 '13 at 2:10

I decided to transform my comment into an answer.

As $x=0$ is not a root, divide the equation by $x^2$ and solve for $x+\frac{1}{x}$ as RGB has pointed out. You will get two solutions $y_1$ and $y_2$ for $x+\frac{1}{x}$: $$x+\frac{1}{x}=\frac{-(a-1)-\sqrt{(a-1)^2+4}}{2}=y_1$$ and $$x+\frac{1}{x}=\frac{-(a-1)+\sqrt{(a-1)^2+4}}{2}=y_2$$ If we want two negative roots the line $y=y_1$ must intercept twice the function $f(x)=x+\frac{1}{x}$ in the third quadrant. That's only possible if $y_1<-2$. See the figure bellow:

enter image description here

So let's solve the inequality: $$y_1<-2 \Rightarrow $$ $$\frac{-(a-1)-\sqrt{(a-1)^2+4}}{2}<-2 \Rightarrow $$ $$-(a-1)-\sqrt{(a-1)^2+4}<-4 \Rightarrow $$ $$(a-1)+\sqrt{(a-1)^2+4}>4 \Rightarrow $$ $$\sqrt{(a-1)^2+4}>5-a \Rightarrow $$ $$a^2-2a+5>25-10a+a^2 \Rightarrow $$ $$8a>20 \Rightarrow $$ $$a>\frac{5}{2}$$ Therefore for $a>\frac{5}{2}$ the original equation will have two negative roots.

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