are roots distinct?

How is it possible to find exactly 1 complex root for a polynomial of real coefficient

Nohow, but root may be double and we may assume as one.

booth wrote: Nohow, but root may be double and we may assume as one. I did not understand what you meant. I too do not understand how a quartic with real coefficients can have exactly 3 real roots? If only one root is complex ( i.e involves $i$), wont it be reflected in the coeffecients? And well the coeffecients are all real since we are not given the option to manipulate them via square roots ( so as to generate some $i$'s). The problem cannot possibly be "about showing how if it had exactly one complex root it would be reflected in the coefficients which isn't possible, Thus no such $a$, exist,"? The question has been falsely copied perhaps?

Assume that $f(x)$ has exactly three real roots. Since it's degree $4$ polynomial it has all roots real, because $ \left(a^2 -2 \right)^2\in\mathbb{R}$. Newton inequalities yield $$\left(\frac{4-2a^2}{4}\right)\ge\frac{6a^2-5}{6}\cdot (a^2-2)^2,\ \left(\frac{6a^2-5}{6}\right)^2\ge\frac{4-2a^2}{4}\cdot 2\iff$$$$(a^2-2)^2(13-12a^2)\ge0,\ \left(a^2-\frac{2+\sqrt{51}}{6}\right)\cdot \left(a^2+\frac{-2+\sqrt{51}}{6}\right)\ge0\iff$$$$(a^2-2)^2(13-12a^2)\ge0,\ a^2\ge\frac{2+\sqrt{51}}{6}\iff$$$$a^2=2.$$We will prove that $a\in\lbrace -\sqrt2,\sqrt2\rbrace$ work. Indeed, we have three real roots $\lbrace 0,0,1+\sqrt{8},1-\sqrt{8}\rbrace$ as $$f(x)=x^2\left((x-1)^2-8\right).\square$$#1712

WolfusA wrote: Indeed, we have three real roots $\lbrace 0,0,1+\sqrt{8},1-\sqrt{8}\rbrace$ That's four real roots. I think both the OP and your solution want to mean three real solutions

Wiz-of-Oz wrote: WolfusA wrote: Indeed, we have three real roots $\lbrace 0,0,1+\sqrt{8},1-\sqrt{8}\rbrace$ That's four real roots. I think both the OP and your solution want to mean three real solutions Yeah 3 "distinct" real roots i guess. But the question says " exactly three real roots"? Maybe the question too meant distinct 3 real ones?

WolfusA Wrong solutions

we can factor the polynomial $\left(-x^2+x\left(1-2a\right)+a^2-2\right)\left(-x^2+x\left(2a+1\right)+a^2-2\right)$ either $\left(-x^2+x\left(1-2a\right)+a^2-2\right)$ or $\left(-x^2+x\left(2a+1\right)+a^2-2\right)$ the discriminant must be 0 $a=\frac{1+\sqrt15}{4},\frac{1-\sqrt15}{4},\frac{-1+\sqrt15}{4},\frac{-1-\sqrt15}{4}$

Jufri wrote: we can factor the polynomial $\left(-x^2+x\left(1-2a\right)+a^2-2\right)\left(-x^2+x\left(2a+1\right)+a^2-2\right)$ either $\left(-x^2+x\left(1-2a\right)+a^2-2\right)$ or $\left(-x^2+x\left(2a+1\right)+a^2-2\right)$ the discriminant must be 0 $a=\frac{1+\sqrt15}{4},\frac{1-\sqrt15}{4},\frac{-1+\sqrt15}{4},\frac{-1-\sqrt15}{4}$ I don't think you can factor it that way, because the given polynomial minus yours gives 6ax^2-6a^2x^2, when it should be 0 if the polynomials were equivalent. Also, I tried the obvious a=$\pm\sqrt{2}$, since a^2-2 would then give 0, and 2a^2-4=0, which meant the polynomial would be of the form x^4-2x^3+(c)x^2, which meant there would be double roots at x=0, or only three distinct roots, but how do I know this is the only answer? Is there a way to get repeated roots (i.e. the polynomial, once given a, can be factored as (x-b)(x-c)(x-d)^2) with a different value of a?

huashiliao2020 wrote: Jufri wrote: we can factor the polynomial $\left(-x^2+x\left(1-2a\right)+a^2-2\right)\left(-x^2+x\left(2a+1\right)+a^2-2\right)$ either $\left(-x^2+x\left(1-2a\right)+a^2-2\right)$ or $\left(-x^2+x\left(2a+1\right)+a^2-2\right)$ the discriminant must be 0 $a=\frac{1+\sqrt15}{4},\frac{1-\sqrt15}{4},\frac{-1+\sqrt15}{4},\frac{-1-\sqrt15}{4}$ I don't think you can factor it that way, because the given polynomial minus yours gives 6ax^2-6a^2x^2, when it should be 0 if the polynomials were equivalent. Also, I tried the obvious a=$\pm\sqrt{2}$, since a^2-2 would then give 0, and 2a^2-4=0, which meant the polynomial would be of the form x^4-2x^3+(c)x^2, which meant there would be double roots at x=0, or only three distinct roots, but how do I know this is the only answer? Is there a way to get repeated roots (i.e. the polynomial, once given a, can be factored as (x-b)(x-c)(x-d)^2) with a different value of a? It does indeed factor like that. The roots are $x=\frac{1{\color{blue}\pm}2a{\color{red}\pm}\sqrt{8a^2{\color{blue}\pm}4a-7}}{2}$ For at least three roots, we need $a\in\left(-\infty,\frac{-1-\sqrt{15}}{4}\right]\cup\left[\frac{1+\sqrt{15}}{4},\infty\right)$. Therefore for exactly three distinct real roots, two of the solutions is to have $a=\frac{-1-\sqrt{15}}{4}$ or $a=\frac{1+\sqrt{15}}{4}$ which both give $x\in\left\{\frac{1-\sqrt{15}}{4},\frac{3+\sqrt{15}\pm2\sqrt{2+2\sqrt{15}}}{4}\right\}$. There are four other possibilities for which two roots are equal. For two of them, there is no real solution for $a $. For the other two, you get $a=\pm\sqrt{2}$ which both give $x\in\left\{0,1\pm 2\sqrt{2}\right\}$.

I see. Thanks.