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.
[Math] 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
quadratics
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Best Answer
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:
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.