Number of real roots in polynomial equation

In summary, a polynomial equation is an equation that contains one or more variables and only uses operations such as addition, subtraction, and multiplication. The number of real roots in a polynomial equation can be determined using the Fundamental Theorem of Algebra or Descartes' Rule of Signs. A polynomial equation can have more than one real root, but the maximum number of real roots is equal to its degree. Real roots are values that satisfy the equation and are real numbers, while complex roots involve the square root of a negative number and cannot be plotted on the number line. A polynomial equation can also have no real roots and instead have complex roots.
  • #1
juantheron
247
1
Evaluate number of real roots of the equation $$x^6-x^5+x^4-x^3+x^2-x+\frac{2}{5} = 0$$
 
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  • #2
jacks said:
Evaluate number of real roots of the equation $$x^6-x^5+x^4-x^3+x^2-x+\frac{2}{5} = 0$$

My solution:

Let \(\displaystyle f(x)=x^6-x^5+x^4-x^3+x^2-x+\frac{2}{5}\)

And then:

\(\displaystyle f'(x)=6x^5-5x^4+4x^3-3x^2+2x-1\)

\(\displaystyle f''(x)=30x^4-20x^3+12x^2-6x+2\)

\(\displaystyle f'''(x)=120x^3-60x^2+24x-6\)

\(\displaystyle f^{(4)}(x)=360x^2-120x+24\)

The 4th derivative has a negative discriminant, therefore as an upward opening parabolic function, is positive for all $x$. This means the third derivative is strictly increasing and can have only 1 real root. Using a numeric root-finding technique, we find this root is approximated by:

\(\displaystyle x\approx0.342384094858369\)

We then look at:

\(\displaystyle f''(0.342384094858369)=0.961949707437654530>0\)

This means we may conclude that the 2nd derivative is positive for all $x$, and so the first derivative is strictly increasing with only 1 real root, which we find at about:

\(\displaystyle x\approx0.67033204760309682774\)

We then look at:

\(\displaystyle f(0.67033204760309682774)=0.03509389397174151671752\)

And so we conclude that for all real $x$, we have $f(x)>0$, and thus $f$ has no real roots.
 
  • #3
Another way:

If $p(x)=x^6-x^5+x^4-x^3+x^2-x+\dfrac{2}{5}$ then $p(-x)=x^6+x^5+x^4+x^3+x^2+x+\dfrac{2}{5}$ and by the Descartes' Rule of Signs there are no negative roots for $p(x)$.

On the other hand, if $f(x)=a_nx^n+\ldots +a_1x+a_0\in\mathbb{R}[x]$ with $a_n\neq 0$ and $c$ a root of de $f(x)$ then $|c|\leq M$ where $$M=\max\left \{\left(n\left| \frac{a_{i-1}}{a_n}\right|\right)^{1/i}:i=1,\ldots,n\right\}.$$ Now, we can use the Sturm's Theorem on the closed interval $\left[0,\lfloor M\rfloor +1\right].$
 
  • #4
My Solution::

$\bullet\; $ If $x\leq 0\;,$ Then $\displaystyle f(x) = x^6-x^5+x^4-x^3+x^2-x+\frac{2}{5}>0$

$\bullet\; $ If $x\geq 1\;,$ Then $\displaystyle f(x) = x^5(x-1)+x^3(x-1)+x(x-1)+\frac{2}{5}>0$

$\bullet \; 0<x<1\;,$ Let $\displaystyle f(x)-\frac{2}{5} = x^6-x^5+x^4-x^3+x^2-x=x(x-1)(x^4+x^2+1)$

So $\displaystyle f(x)-\frac{2}{5} = -x(1-x)(1+x+x^2)(1-x+x^2) = -(1-x^3)(x-x^2+x^3)$

So $\displaystyle -f(x)+\frac{2}{5} = (1-x^3)(x-x^2+x^3) \leq \left(\frac{1-x^3+x-x^2+x^3}{2}\right)^2$

So $\displaystyle -f(x)+\frac{2}{5}\leq \left(\frac{x-x^2+1}{2}\right)^2 = \frac{1}{64}\left(4x-4x^2+4\right)^2$

So $\displaystyle -f(x)+\frac{2}{5} \leq \frac{1}{64}\left(5-(2x-1)^2\right) \leq \frac{25}{64}$

So $\displaystyle -f(x)+\frac{2}{5}\leq \frac{25}{64}\Rightarrow f(x)-\frac{2}{5}\geq \frac{25}{64}$

So $\displaystyle f(x)\geq \frac{2}{5}-\frac{25}{64} = \frac{3}{320}>0$

So $\displaystyle f(x) = x^6-x^5+x^4-x^3+x^2-x+\frac{2}{5}>0\;\forall x \in \mathbb{R}$
 
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1. What is a polynomial equation?

A polynomial equation is an equation that contains one or more variables and only uses operations such as addition, subtraction, and multiplication. The variables in a polynomial equation can only have integer exponents, and the coefficients must be real numbers.

2. How do you determine the number of real roots in a polynomial equation?

To determine the number of real roots in a polynomial equation, you can use the Fundamental Theorem of Algebra, which states that a polynomial of degree n has exactly n complex roots (including real and imaginary). You can also use the Descartes' Rule of Signs to estimate the number of positive and negative real roots in a polynomial equation.

3. Can a polynomial equation have more than one real root?

Yes, a polynomial equation can have more than one real root. For example, the equation x^2 - 2x + 1 = 0 has two real roots, x = 1 and x = 1. However, the maximum number of real roots a polynomial equation can have is equal to its degree. So, a polynomial equation of degree 3 can have a maximum of 3 real roots.

4. What is the difference between real roots and complex roots?

Real roots are values of the variable that satisfy the polynomial equation and are real numbers, meaning they are on the number line. Complex roots are also values that satisfy the polynomial equation, but they are imaginary numbers, meaning they involve the square root of a negative number and cannot be plotted on the number line.

5. Can a polynomial equation have no real roots?

Yes, a polynomial equation can have no real roots. For example, the equation x^2 + 1 = 0 has no real roots because there is no real number that, when squared, will give a result of -1. In this case, the equation has two complex roots, x = i and x = -i, where i is the imaginary unit.

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