Find the sum of the real roots of the equation $6x^4+9x^3-15x^2+9x+6=0$

Question

Find the sum of the real roots of the equation $$6x^4+9x^3-15x^2+9x+6=0$$

We can solve the equation by dividing both sides by $x^2\ne0$. Then we'll get $$6\left(x^2+\dfrac{1}{x^2}\right)+9\left(x+\dfrac{1}{x}\right)-15=0$$ which by putting $t=x+\dfrac{1}{x}$ can be written as $$6(t^2-2)+9t-15=0\iff 6t^2+9t-27=0$$

We can find the roots from here. This just seems to me like a lot of work to simply find the sum.

The sum we're searching for is $-3$: . See this

I would like to ask if there is a faster approach? Does needing only the sum of the real roots make things more complicated for finding a faster approach$?$

Answer 1

Well, if $\alpha$ is a root of the palindromic polynomial, so is $\frac1\alpha$, so the real (and non-real) roots occur in inverse pairs. Thus when you use the substitution $t=x+\frac1x$, each solution for $t$ is already the sum of a real root pair (or a non-real, conjugate root pair).

Hence once you have the equation $6t^2+9t-27=0 \iff (2t-3)(t+3)=0$, you actually already have each sum of paired inverse roots. $t=\frac32$ represents a non-real pair (as $|t|=|x+\frac1x|\geqslant 2$ for reals), while $t=-3$ represents the sum of a real pair of roots. As there is only one root with $|t|\geqslant 2$, you have the sum of real roots as $-3$.

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P.S. You may want to try showing rigorously that $|t|<2$ iff the corresponding $x,\frac1x$ are non-real.

Answer 2

Let $$P(x)=2x^4+3x^3-5x^2+3x+2.$$

The sum of complex roots of $P$ is $-\frac32.$

Let us now concentrate on your less obvious question of the sum of real roots with a "faster approach" than your change of variable "$t=x+\frac1x$". Note however that you were very close to the goal since, as noted by @dxiv in comment, $$6t^2+9t-27 = 3 (2 t - 3) (t + 3) $$$$= \frac{3}{x^2} (2 x^2 - 3 x + 2) (x^2 + 3 x + 1)$$ and this factorization readily gave you the answer, as we shall see below.

But you can avoid all these intermediate steps, writing instead directly $$P(x)=2(x^2-ax+1)(x^2-bx+1)$$ where $$-2(a+b)=3\quad\text{and}\quad2(ab+2)=-5,$$ i.e. $$P(x)=2(x^2+3x+1)\left(x^2-\frac32x+1\right).$$ Since the discriminant of the first quadratic factor is positive and the other one is negative, the real roots of $P$ are the two roots of the first factor. Their sum is $-3.$

Answer 3

Upon recognizing that the quartic polynomial is palindromic, a "fast" approach might be one where we could find the sum of real roots (if they exist) directly from the (real) coefficients. [The theory for this approach is related to the arguments given by Macavity and Anne Bauval.]

Since the zeroes are linked in pairs by equations $ \ x + \frac{1}{x} \ = \ \alpha \ \ , \ \ x + \frac{1}{x} \ = \ \beta \ \ , \ $ the polynomial is factorized by $$ a·(x^2 \ - \ \alpha·x \ + \ 1)·(x^2 \ - \ \beta·x \ + \ 1) $$ $$ = \ \ a·(x^4 \ - \ [ \ \alpha \ + \ \beta \ ]·x^3 \ + \ [ \ \alpha · \beta \ + \ 2 \ ]·x^2 \ - \ [ \ \alpha \ + \ \beta \ ]·x \ + \ 1) \ \ . $$ Setting this against the polynomial $ \ ax^4 + bx^3 + cx^2 + bx + a \ \ $ leads to $$ \alpha \ + \ \beta \ \ = \ -\frac{b}{a} \ \ \ , \ \ \ \alpha·\beta \ \ = \ \ \frac{c}{a} \ - \ 2 \ \ \Rightarrow \ \ (\alpha \ + \ \beta)^2 \ \ = \ \ \frac{b^2}{a^2}\ \ \ , \ \ \ 4·\alpha·\beta \ \ = \ \ \frac{4c - 8a}{a} $$ $$ \Rightarrow \ \ (\alpha \ - \ \beta)^2 \ \ = \ \ \frac{8a^2 \ + \ b^2 \ - \ 4ac}{a^2} $$ [so this looks like we are deriving the quadratic formula] $$ \alpha \ , \ \beta \ \ = \ \ \frac{-b \ \pm \ \sqrt{ \ 8a^2 \ + \ b^2 \ - \ 4ac} }{2a} \ \ . $$ The difference in how we apply this from what we often do is that we don't want to look at the discriminant, but at the values of $ \ \alpha \ $ and $ \ \beta \ \ . \ $ The cases are

• for $ \ |\alpha| \ \ge \ 2 \ $ and $ \ |\beta| \ \ge \ 2 \ \ , \ $ the sum of the real zeroes is $ \ \alpha \ + \ \beta \ \ = \ -\frac{b}{a} \ \ , \ $ which is simply what the Viete relation tells us;

• for $ \ |\alpha| \ \ge \ 2 \ $ only or $ \ |\beta| \ \ge \ 2 \ $ only, the sum of the real zeroes is $ \ \alpha \ $ or $ \ \beta \ \ , \ $ respectively;

• for $ \ |\alpha| \ < \ 2 \ $ and $ \ |\beta| \ < \ 2 \ \ , \ $ there are no real zeroes, so the sum of the real zeroes is $ \ 0 \ \ . $

Some illustrative examples are:

$ x^4 \ - \ 4x^3 \ + \ 6x^2 \ - \ 4x \ + \ 1 \ \ = \ \ (x - 1)^4 \ \ : \ \ \alpha \ , \ \beta \ \ = \ \ \frac{4 \ \pm \ \sqrt{ \ 8 \ + \ 16 \ - \ 24} }{2} \ \ = \ \ 2 \ \ , \ \ 2 \ \ ; $ $ \quad \quad \quad \quad \text{sum of real zeroes:} \ \ 2 + 2 \ = \ 4 \ \ \ ; \ \ \text{sum of all zeroes:} \ \ 4 $

$ x^4 \ - \ 2x^2 \ + \ 1 \ \ = \ \ (x - 1)^2·(x + 1)^2 \ \ : \ \ \alpha \ , \ \beta \ \ = \ \ \frac{0 \ \pm \ \sqrt{ \ 8 \ + \ 0 \ + \ 8} }{2} \ \ = \ \ 2 \ \ , \ \ -2 \ \ ; $ $ \quad \quad \quad \quad \text{sum of real zeroes:} \ \ 2 + (-2) \ = \ 0 \ \ \ ; \ \ \text{sum of all zeroes:} \ \ 0 $

$ 3x^4 \ + \ 6x^3 \ - \ 39x^2 \ + \ 6x \ + \ 3 \ \ = \ \ 3·(x^2 - 3x + 1)·(x^2 + 5x + 1) \ \ : $ $ \quad \quad \quad \quad \alpha \ , \ \beta \ \ = \ \ \frac{-6 \ \pm \ \sqrt{ \ 72 \ + \ 36 \ + \ 468} }{6} \ \ = \ \ 3 \ \ , \ \ -5 \ \ ; $ $ \quad \quad \quad \quad \text{sum of real zeroes:} \ \ 3 + (-5) \ = \ -2 \ \ \ ; \ \ \text{sum of all zeroes:} \ \ -2 $

$ \mathbf{6x^4 \ + \ 9x^3 \ - \ 15x^2 \ + \ 9x \ + \ 6 } \ \ = \ \ 6·\left(x^2 - \frac32 x + 1 \right)·(x^2 + 3x + 1) \ \ : $ $ \quad \quad \quad \quad \alpha \ , \ \beta \ \ = \ \ \frac{-9 \ \pm \ \sqrt{ \ 288 \ + \ 81 \ + \ 360} }{12} \ \ = \ \ [\frac32] \ \ , \ \ -3 \ \ ; \ \ \ \text{sum of real zeroes:} \ \ -3 \ \ \ ; $ $ \quad \quad \quad \quad \text{sum of all zeroes:} \ \ -\frac32 $

$ 9x^4 \ + \ 3x^3 \ + \ 16x^2 \ + \ 3x \ + \ 9 \ \ = \ \ 9·\left(x^2 - \frac13 x + 1 \right)·(x^2 + \frac23 x + 1) \ \ : $ $ \quad \quad \quad \quad \alpha \ , \ \beta \ \ = \ \ \frac{-3 \ \pm \ \sqrt{ \ 648 \ + \ 9 \ - \ 576} }{18} \ \ = \ \ [\frac13] \ \ , \ \ [-\frac23] \ \ ; \ \ \ \text{sum of real zeroes:} \ \ 0 \ \ \ ; $ $ \quad \quad \quad \quad \text{sum of all zeroes:} \ \ -\frac13 \ \ . $

This demonstrates the limitation of applying the Viete relations alone and indicates how the non-real zeroes can be "sieved out" from the sum of real zeroes. However, this method is not all that much faster than the approach you are taking; also, since it is specialized to this particular sort of polynomial (palindromic quartic), it is not something really worth committing to memory.