Find the number of real roots of the equation $54x^4-36x^3+18x^2-6x+1=0$

Question

I entered the equation in desmos.com and no roots were lying below or x=0 lines , hence all roots are imaginary.

Using Descartes rule for f(x) 4 sign change occurs , hence 4,2 or zero positive roots.

Using Descartes rule for f(-x) 0 sign change occurs , hence zero negative roots.

For here on how do I proceed

In general Sturm's theorem gives you an algorithm to do that. — user647486, Apr 28 '19 at 06:01
derive and solve the cubic you find all local minima/maxima are positive, hence no real roots. — abc..., Apr 28 '19 at 06:02

Toby Mak · Accepted Answer · 2019-04-28T06:50:37.653

2

To expand on Dr. Sonnhard Graubner's answer, let $f(x) = 54x^4-36x^2+15x^2-6x+1$. Then:

$$54x^4-36x^3+18x^2-6x+1 > f(x)$$ $$\Rightarrow (3x-1)^2(6x^2+1)$$

by the rational root test.

The smaller function has only one root: $x = \frac{1}{3}$, and there $f''(x) = 648x^2-216x+30$ is positive. Therefore, for all real $x$, $f(x) > 0$ and therefore $54x^4-36x^3+18x^2-6x+1 > 0$.

edited Apr 28 '19 at 06:50

answered Apr 28 '19 at 06:43

Toby Mak

16,827

How did I miss that: $(3x - 1)^2 ≥ 0$ and $6x^2 + 1 > 0$ for all real $x$, so their product must be positive considering what happens at $x = \frac{1}{3}$, the only case we need to check. – Toby Mak Aug 07 '22 at 06:22

score 1 · Answer 2 · answered Apr 28 '19 at 06:18

1

Hint: Prove that $$54x^4-36x^3+18x^2-6x+1>0$$ for all real $$x$$

answered Apr 28 '19 at 06:18

Dr. Sonnhard Graubner

95,283

Find the number of real roots of the equation $54x^4-36x^3+18x^2-6x+1=0$

2 Answers2