The equation $x^4-2x^3-3x^2+4x-1=0$ has four distinct real roots $x_1,x_2,x_3,x_4$ such that $x_1<x_2<x_3<x_4$ and product of two roots is unity.

Question

The equation $x^4-2x^3-3x^2+4x-1=0$ has four distinct real roots $x_1,x_2,x_3,x_4$ such that $x_1<x_2<x_3<x_4$ and product of two roots is unity, then:

$Q-1$: Find $x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4$

$Q-2$: Find $x_2^3+x_4^3$

My attempt is as follows:-

$A-1$ : First I tried to find any trivial root, but was not able to find any. After that I tried following:-

$$x_1\cdot x_2+x_1\cdot x_3+x_1\cdot x_4+x_2\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3$$ $$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3-x_1\cdot x_4-x_2\cdot x_3$$

$$x_1\cdot x_2\cdot x_3\cdot x_4=-1$$ $$x_1\cdot x_4=\dfrac{-1}{x_2\cdot x_3}$$

$$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3-x_1\cdot x_4-x_2\cdot x_3$$ $$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3-x_2\cdot x_3+\dfrac{1}{x_2\cdot x_3}$$

But from here I was not able to proceed as I was not able to calculate value of $x_2\cdot x_3$

$A-2$ : $(x_2+x_4)(x_2^2+x_4^2-x_2\cdot x_4)$

Now here I was not getting any idea for how to proceed.

Please help me in this.

Answer 1

Hint The product of two roots is $-1$ and the product of the other two roots is $1$.

Therefore $$x^4-2x^3-3x^2+4x-1=(x^2+ax+1)(x^2+bx-1)$$

Oppening the brackets gives $$a+b=-2\\ ab=-3 \\ b-a=4$$ which is trivial to solve.

Answer 2

Doing it in OP's way

$f(x)=x^4-2x^3-3x^2+4x-1=0$ let its rootsa be $a,b,c,d$, and let $a+b=u$ and $ab=v.$ Then by Vieta's formulas: $$a+b+c+d=2~~~(1) \implies c+d=2-u$$ $$abcd=-1 ~~~~~(2) \implies cd=-1/v$$ $$ab+bc+cd+ac+bd+ad=-3~~~(3) \implies v-1/v+(a+b)(c+d)=-3 \implies v-1/v+u(2-u)=-3$$ $$abc+bcd+acd+bcd=-4~~~(4) \implies ab(c+d)+cd(a+b) =-4 \implies v(2-u)-(1/v)u=-4$$By putting $v=1$ in (3) we get $u^2-2u-3=0 \implies u=3,-1$ Next $a+b=3, ab=1; a+b=-1,ab=1$ give $$a, b=\frac{3\pm \sqrt{5}}{2};~~ a,b=\frac{-1\pm \sqrt{5}}{2}$$ These for are the roots which can ve arranged ascending order as $$x=\frac{-1-\sqrt{5}}{2},\frac{3-\sqrt{5}}{2},\frac{-1+\sqrt{5}}{2},\frac{3+\sqrt{5}}{2}~~~~(5)$$ Interestingly, (4) when $v=1$ also gives $u=3$, again.

Answer 3

Also, we can use the following way.

For any value of $k$ we obtain: $$x^4-2x^3-3x^2+4x-1=(x^2-x+k)^2-x^2-k^2+2kx-2kx^2-3x^2+4x-1=$$ $$=(x^2-x+k)^2-((2k+4)x^2-(2k+4)x+k^2+1),$$ which for $k=0$ gives: $$x^4-2x^3-3x^2+4x-1=(x^2-x)^2-(2x-1)^2=(x^2-3x+1)(x^2+x-1).$$ Can you end it now?