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How many real roots does the equation $x^5 - 5x + 2 =0$ have?

I know the following facts:

  1. The equation will have odd number of real root.
  2. That function cannot have rational root.
  3. The function will have two real roots between $(1,2)$ and $(0,1)$.

Can anyone please help me in solving this problem?

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    $\begingroup$ So, do you know Rolle's Theorem? Or do you know Descartes' Rule of Sign? $\endgroup$ – GEdgar Jul 20 '18 at 12:06
  • $\begingroup$ There is another real root in $(-2,-1)$. $\endgroup$ – Batominovski Jul 20 '18 at 12:10
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Hint. Note that $f'(x)=5(x^4-1)=5(x^2+1)(x-1)(x+1)$ which implies that $f$ is strictly increasing in $(-\infty,-1]$, it is strictly decreasing in $[-1,1]$ and it is strictly increasing in $[1,+\infty)$. Knowing that $\lim_{x\to \pm \infty}f(x)=\pm\infty$, $f(-1)=6$, and $f(1)=-2$, what may we conclude?

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  • $\begingroup$ But how can it have 3 real roots? Because every root will be irrational or integer and I can understand it can not have any integer root. Then how it can have three irrational root when we know that irrational roots will be there in conjugates.?@Robert Z $\endgroup$ – cmi Jul 20 '18 at 12:47
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    $\begingroup$ @cmi complex roots come in conjugate pairs not real roots (irrationals or rationals) $\endgroup$ – Robert Z Jul 20 '18 at 13:41
  • $\begingroup$ No if $a + {b^(1/2)}$is a root of the equation with rational coefficients then $a - {b^(1/2)}$ will be it's root as well. $\endgroup$ – cmi Jul 20 '18 at 15:26
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    $\begingroup$ If $t$ is an irrational real root of such polynomial then it does not follow that $t$ has the form $a+\sqrt{b}$ $\endgroup$ – Robert Z Jul 20 '18 at 15:35
  • $\begingroup$ Now I am clear. Thank you. $\endgroup$ – cmi Jul 20 '18 at 15:38
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Hint:

If you take the derivative twice, you get $x\mapsto20x^3$, which is negative for $x<0$ and positive for $x>0$. This tells you that $f'$ is decreasing on $\Bbb R_-$ and increasing on $\Bbb R_+$. Therefore $f'$ can vanish at most twice.

What would happen to $f'$ if $f(x)=0$ had $5$ solutions?

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Using Descartes' Rule of Signs ...

$p(x) = x^5-5x+2$ has two sign differences, so $p(x)$ has either $2$ or $0$ positive zeros.

$p(-y) = -y^5 + 5y + 2$ has one sign difference, so $p(-y)$ has $1$ positive zero, that is $p(x)$ has $1$ negative zero.

From your fact 3, we conclude $p(x)$ has at least $2$ positive zeros.

Result: $p(x)$ has exactly $3$ zeros, two of them positive, one of them negative.

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We can use Sturm's theorem to find a definitive answer to this (unlike Descartes, it counts exactly how many distinct real roots there are). A Sturm chain for $X^5-5X+2$ is given by $$ \left( X^5-5X+2 , 5X^4 -5 , 4X-2 , \frac{75}{16} \right). $$ The last term is a multiple of the discriminant, and in particular, is not zero, so there are no repeated roots. It suffices to examine the sign changes between the leading coefficients, and subtract from the number of sign changes in the leading coefficients when $X$ is replaced by $-X$. The former chain is $$ (1,5,4,75/16), $$ which has no sign changes, while the latter is $$ (-1,5,-4,75/16), $$ which has three sign changes. Hence there are $3-0=3$ real roots.

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  • $\begingroup$ No if $a + {b^(1/2)}$is a root of the equation with rational coefficients then $a - {b^(1/2)}$ will be it's root as well. $\endgroup$ – cmi Jul 20 '18 at 15:31
  • $\begingroup$ But how can it have 3 real roots? Because every real root will be irrational or integer and I can understand it can not have any integer root. Then how it can have three irrational root when we know that irrational roots will be there in conjugates.?Batominovski $\endgroup$ – cmi Jul 20 '18 at 15:31
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As you have worked out, there is at least one root in each of the intervals $(0,1)$ and $(1,2)$. The number of positive roots as given by the rule of signs is zero or two, so there are exactly two positive roots. Applying the rule of signs to the polynomial with $x$ replaced by $-x$ ($-x^5+5x+2$) we see there is exactly one negative root. Lastly, $x=0$ is obviously not a root, so there are exactly three real roots of the polynomial.

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Here is a solution without calculus or Descartes's Rule of Signs. However, some knowledge about continuity of polynomial functions is required.

Let $a,b,c,d,e$ be the roots of this polynomial. Using Vieta's Formulas, we have $$a+b+c+d+e=0$$ and $$ab+ac+ad+ae+bc+bd+be+cd+ce+de=0\,.$$ This means $$\begin{align} a^2&+b^2+c^2+d^2+e^2 \\&=(a+b+c+d+e)^2-2(ab+ac+ad+ae+bc+bd+be+cd+ce+de) \\&=0-2\cdot 0=0\,.\end{align}$$ Since $0$ is not a root of this polynomial, we conclude that not all roots are real (otherwise, it must hold that $a^2+b^2+c^2+d^2+e^2>0$). Thus, the polynomial has either one or three real roots. Since the polynomial has at least one root in each of the three intervals $(-2,-1)$, $(0,1)$, and $(1,2)$, we conclude that there are exactly three real roots.

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  • $\begingroup$ But how can it have 3 real roots? Because every root will be irrational or integer and I can understand it can not have any integer root. Then how it can have three irrational root when we know that irrational roots will be there in conjugates.?Batominovski $\endgroup$ – cmi Jul 20 '18 at 12:46
  • $\begingroup$ I don't understand your question. This problem is about real roots. Why are you asking about irrational roots? What does "conjugate irrational numbers" mean in your case? $\endgroup$ – Batominovski Jul 20 '18 at 12:50
  • $\begingroup$ I am not asking about irrational roots. I am giving an argument. Can you please read my answer one more time?@Batominovski $\endgroup$ – cmi Jul 20 '18 at 12:52
  • $\begingroup$ I replied. What does it mean for irrational numbers to be conjugates? This is not a quadratic polynomial. I think you are very confused. $\endgroup$ – Batominovski Jul 20 '18 at 12:53
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    $\begingroup$ Sure, IF $a+\sqrt{b}$, where $a$ is a rational and $b$ is a nonsquare rational, is a root of a rational polynomial, you do get that $a-\sqrt{b}$ is also a root. But not all irrational roots take this form. I thought Abcd already explained to you. Do you care to read the link Abcd gave you? It's quite exhausting to explain to somebody who does not take the explanation into consideration. $\endgroup$ – Batominovski Jul 20 '18 at 15:48
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Just adding to RobertZ's precise answer:

enter image description here

Beyond $1$ and before $-1$, the function is strictly increasing.

Since $f(-1)= 6$ and $f(1)= -2$, from Intermediate value theorem, the function must attain a zero in $(-1,1)$. Similarly it attains a zero in $(1, \infty)$ and $(-\infty , -1)$.

So it has $3$ roots.

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  • $\begingroup$ But how can it have 3 real roots? Because every root will be irrational or integer and I can understand it can not have any integer root. Then how it can have three irrational root when we know that irrational roots will be there in conjugates.? $\endgroup$ – cmi Jul 20 '18 at 12:45
  • $\begingroup$ @cmi check this: math.stackexchange.com/questions/2608475/… $\endgroup$ – Archer Jul 20 '18 at 12:55

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