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The number of real roots of $x^5 - 5x + 2 =0$
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How many real roots does the equation $x^5 - 5x + 2 =0$ have?
I know the following facts:
- The equation will have odd number of real root.
- That function cannot have rational root.
- The function will have two real roots between $(1,2)$ and $(0,1)$.
Can anyone please help me in solving this problem?
algebra-precalculus polynomials roots
asked Jul 20 at 12:00
cmi
64510
add a comment |Â
up vote
3
down vote
favorite
How many real roots does the equation $x^5 - 5x + 2 =0$ have?
I know the following facts:
- The equation will have odd number of real root.
- That function cannot have rational root.
- The function will have two real roots between $(1,2)$ and $(0,1)$.
Can anyone please help me in solving this problem?
algebra-precalculus polynomials roots
asked Jul 20 at 12:00
cmi
64510
1
So, do you know Rolle's Theorem? Or do you know Descartes' Rule of Sign?
– GEdgar
Jul 20 at 12:06
There is another real root in $(-2,-1)$.
– Batominovski
Jul 20 at 12:10
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
How many real roots does the equation $x^5 - 5x + 2 =0$ have?
I know the following facts:
- The equation will have odd number of real root.
- That function cannot have rational root.
- The function will have two real roots between $(1,2)$ and $(0,1)$.
Can anyone please help me in solving this problem?
algebra-precalculus polynomials roots
asked Jul 20 at 12:00
cmi
64510
How many real roots does the equation $x^5 - 5x + 2 =0$ have?
I know the following facts:
- The equation will have odd number of real root.
- That function cannot have rational root.
- The function will have two real roots between $(1,2)$ and $(0,1)$.
Can anyone please help me in solving this problem?
algebra-precalculus polynomials roots
asked Jul 20 at 12:00
cmi
64510
edited Jul 20 at 14:40
asked Jul 20 at 12:00
cmi
64510
asked Jul 20 at 12:00
cmi
64510
asked Jul 20 at 12:00
cmi
64510
64510
1
So, do you know Rolle's Theorem? Or do you know Descartes' Rule of Sign?
– GEdgar
Jul 20 at 12:06
There is another real root in $(-2,-1)$.
– Batominovski
Jul 20 at 12:10
add a comment |Â
1
So, do you know Rolle's Theorem? Or do you know Descartes' Rule of Sign?
– GEdgar
Jul 20 at 12:06
There is another real root in $(-2,-1)$.
– Batominovski
Jul 20 at 12:10
1
1
So, do you know Rolle's Theorem? Or do you know Descartes' Rule of Sign?
– GEdgar
Jul 20 at 12:06
So, do you know Rolle's Theorem? Or do you know Descartes' Rule of Sign?
– GEdgar
Jul 20 at 12:06
There is another real root in $(-2,-1)$.
– Batominovski
Jul 20 at 12:10
There is another real root in $(-2,-1)$.
– Batominovski
Jul 20 at 12:10
add a comment |Â
7 Answers
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7
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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 $lim_xto pm inftyf(x)=pminfty$, $f(-1)=6$, and $f(1)=-2$, what may we conclude?
answered Jul 20 at 12:06
Robert Z
84k954122
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
– cmi
Jul 20 at 12:47
1
@cmi complex roots come in conjugate pairs not real roots (irrationals or rationals)
– Robert Z
Jul 20 at 13:41
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.
– cmi
Jul 20 at 15:26
2
If $t$ is an irrational real root of such polynomial then it does not follow that $t$ has the form $a+sqrtb$
– Robert Z
Jul 20 at 15:35
Now I am clear. Thank you.
– cmi
Jul 20 at 15:38
add a comment |Â
up vote
2
down vote
Hint:
If you take the derivative twice, you get $xmapsto20x^3$, 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?
answered Jul 20 at 12:08
Arnaud Mortier
19k22159
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up vote
2
down vote
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.
answered Jul 20 at 12:12
GEdgar
58.4k264163
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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.
answered Jul 20 at 12:08
Parcly Taxel
33.6k136588
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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
$$beginalign a^2&+b^2+c^2+d^2+e^2
\&=(a+b+c+d+e)^2-2(ab+ac+ad+ae+bc+bd+be+cd+ce+de)
\&=0-2cdot 0=0,.endalign$$
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.
answered Jul 20 at 12:18
Batominovski
23.2k22777
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
– cmi
Jul 20 at 12:46
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?
– Batominovski
Jul 20 at 12:50
I am not asking about irrational roots. I am giving an argument. Can you please read my answer one more time?@Batominovski
– cmi
Jul 20 at 12:52
I replied. What does it mean for irrational numbers to be conjugates? This is not a quadratic polynomial. I think you are very confused.
– Batominovski
Jul 20 at 12:53
1
Sure, IF $a+sqrtb$, where $a$ is a rational and $b$ is a nonsquare rational, is a root of a rational polynomial, you do get that $a-sqrtb$ 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.
– Batominovski
Jul 20 at 15:48
 |Â
show 1 more comment
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1
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Just adding to RobertZ's precise answer:
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.
answered Jul 20 at 12:36
Abcd
2,3761624
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.?
– cmi
Jul 20 at 12:45
@cmi Irrational roots need come in pairs
– Abcd
Jul 20 at 12:55
@cmi check this: math.stackexchange.com/questions/2608475/…
– Abcd
Jul 20 at 12:55
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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
$$ left( X^5-5X+2 , 5X^4 -5 , 4X-2 , frac7516 right). $$
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.
answered Jul 20 at 13:00
Chappers
55k74190
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.
– cmi
Jul 20 at 15:31
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
– cmi
Jul 20 at 15:31
add a comment |Â
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
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 $lim_xto pm inftyf(x)=pminfty$, $f(-1)=6$, and $f(1)=-2$, what may we conclude?
answered Jul 20 at 12:06
Robert Z
84k954122
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
– cmi
Jul 20 at 12:47
1
@cmi complex roots come in conjugate pairs not real roots (irrationals or rationals)
– Robert Z
Jul 20 at 13:41
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.
– cmi
Jul 20 at 15:26
2
If $t$ is an irrational real root of such polynomial then it does not follow that $t$ has the form $a+sqrtb$
– Robert Z
Jul 20 at 15:35
Now I am clear. Thank you.
– cmi
Jul 20 at 15:38
add a comment |Â
up vote
7
down vote
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 $lim_xto pm inftyf(x)=pminfty$, $f(-1)=6$, and $f(1)=-2$, what may we conclude?
answered Jul 20 at 12:06
Robert Z
84k954122
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
– cmi
Jul 20 at 12:47
1
@cmi complex roots come in conjugate pairs not real roots (irrationals or rationals)
– Robert Z
Jul 20 at 13:41
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.
– cmi
Jul 20 at 15:26
2
If $t$ is an irrational real root of such polynomial then it does not follow that $t$ has the form $a+sqrtb$
– Robert Z
Jul 20 at 15:35
Now I am clear. Thank you.
– cmi
Jul 20 at 15:38
add a comment |Â
up vote
7
down vote
up vote
7
down vote
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 $lim_xto pm inftyf(x)=pminfty$, $f(-1)=6$, and $f(1)=-2$, what may we conclude?
answered Jul 20 at 12:06
Robert Z
84k954122
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 $lim_xto pm inftyf(x)=pminfty$, $f(-1)=6$, and $f(1)=-2$, what may we conclude?
answered Jul 20 at 12:06
Robert Z
84k954122
edited Jul 20 at 12:13
answered Jul 20 at 12:06
Robert Z
84k954122
answered Jul 20 at 12:06
Robert Z
84k954122
answered Jul 20 at 12:06
Robert Z
84k954122
84k954122
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
– cmi
Jul 20 at 12:47
1
@cmi complex roots come in conjugate pairs not real roots (irrationals or rationals)
– Robert Z
Jul 20 at 13:41
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.
– cmi
Jul 20 at 15:26
2
If $t$ is an irrational real root of such polynomial then it does not follow that $t$ has the form $a+sqrtb$
– Robert Z
Jul 20 at 15:35
Now I am clear. Thank you.
– cmi
Jul 20 at 15:38
add a comment |Â
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
– cmi
Jul 20 at 12:47
1
@cmi complex roots come in conjugate pairs not real roots (irrationals or rationals)
– Robert Z
Jul 20 at 13:41
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.
– cmi
Jul 20 at 15:26
2
If $t$ is an irrational real root of such polynomial then it does not follow that $t$ has the form $a+sqrtb$
– Robert Z
Jul 20 at 15:35
Now I am clear. Thank you.
– cmi
Jul 20 at 15:38
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
– cmi
Jul 20 at 12:47
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
– cmi
Jul 20 at 12:47
1
1
@cmi complex roots come in conjugate pairs not real roots (irrationals or rationals)
– Robert Z
Jul 20 at 13:41
@cmi complex roots come in conjugate pairs not real roots (irrationals or rationals)
– Robert Z
Jul 20 at 13:41
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.
– cmi
Jul 20 at 15:26
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.
– cmi
Jul 20 at 15:26
2
2
If $t$ is an irrational real root of such polynomial then it does not follow that $t$ has the form $a+sqrtb$
– Robert Z
Jul 20 at 15:35
If $t$ is an irrational real root of such polynomial then it does not follow that $t$ has the form $a+sqrtb$
– Robert Z
Jul 20 at 15:35
Now I am clear. Thank you.
– cmi
Jul 20 at 15:38
Now I am clear. Thank you.
– cmi
Jul 20 at 15:38
add a comment |Â
up vote
2
down vote
Hint:
If you take the derivative twice, you get $xmapsto20x^3$, 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?
answered Jul 20 at 12:08
Arnaud Mortier
19k22159
add a comment |Â
up vote
2
down vote
Hint:
If you take the derivative twice, you get $xmapsto20x^3$, 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?
answered Jul 20 at 12:08
Arnaud Mortier
19k22159
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint:
If you take the derivative twice, you get $xmapsto20x^3$, 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?
answered Jul 20 at 12:08
Arnaud Mortier
19k22159
Hint:
If you take the derivative twice, you get $xmapsto20x^3$, 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?
answered Jul 20 at 12:08
Arnaud Mortier
19k22159
answered Jul 20 at 12:08
Arnaud Mortier
19k22159
answered Jul 20 at 12:08
Arnaud Mortier
19k22159
answered Jul 20 at 12:08
Arnaud Mortier
19k22159
19k22159
add a comment |Â
add a comment |Â
up vote
2
down vote
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.
answered Jul 20 at 12:12
GEdgar
58.4k264163
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up vote
2
down vote
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.
answered Jul 20 at 12:12
GEdgar
58.4k264163
add a comment |Â
up vote
2
down vote
up vote
2
down vote
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.
answered Jul 20 at 12:12
GEdgar
58.4k264163
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.
answered Jul 20 at 12:12
GEdgar
58.4k264163
answered Jul 20 at 12:12
GEdgar
58.4k264163
answered Jul 20 at 12:12
GEdgar
58.4k264163
answered Jul 20 at 12:12
GEdgar
58.4k264163
58.4k264163
add a comment |Â
add a comment |Â
up vote
1
down vote
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.
answered Jul 20 at 12:08
Parcly Taxel
33.6k136588
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up vote
1
down vote
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.
answered Jul 20 at 12:08
Parcly Taxel
33.6k136588
add a comment |Â
up vote
1
down vote
up vote
1
down vote
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.
answered Jul 20 at 12:08
Parcly Taxel
33.6k136588
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.
answered Jul 20 at 12:08
Parcly Taxel
33.6k136588
answered Jul 20 at 12:08
Parcly Taxel
33.6k136588
answered Jul 20 at 12:08
Parcly Taxel
33.6k136588
answered Jul 20 at 12:08
Parcly Taxel
33.6k136588
33.6k136588
add a comment |Â
add a comment |Â
up vote
1
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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
$$beginalign a^2&+b^2+c^2+d^2+e^2
\&=(a+b+c+d+e)^2-2(ab+ac+ad+ae+bc+bd+be+cd+ce+de)
\&=0-2cdot 0=0,.endalign$$
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.
answered Jul 20 at 12:18
Batominovski
23.2k22777
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
– cmi
Jul 20 at 12:46
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?
– Batominovski
Jul 20 at 12:50
I am not asking about irrational roots. I am giving an argument. Can you please read my answer one more time?@Batominovski
– cmi
Jul 20 at 12:52
I replied. What does it mean for irrational numbers to be conjugates? This is not a quadratic polynomial. I think you are very confused.
– Batominovski
Jul 20 at 12:53
1
Sure, IF $a+sqrtb$, where $a$ is a rational and $b$ is a nonsquare rational, is a root of a rational polynomial, you do get that $a-sqrtb$ 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.
– Batominovski
Jul 20 at 15:48
 |Â
show 1 more comment
up vote
1
down vote
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
$$beginalign a^2&+b^2+c^2+d^2+e^2
\&=(a+b+c+d+e)^2-2(ab+ac+ad+ae+bc+bd+be+cd+ce+de)
\&=0-2cdot 0=0,.endalign$$
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.
answered Jul 20 at 12:18
Batominovski
23.2k22777
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
– cmi
Jul 20 at 12:46
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?
– Batominovski
Jul 20 at 12:50
I am not asking about irrational roots. I am giving an argument. Can you please read my answer one more time?@Batominovski
– cmi
Jul 20 at 12:52
I replied. What does it mean for irrational numbers to be conjugates? This is not a quadratic polynomial. I think you are very confused.
– Batominovski
Jul 20 at 12:53
1
Sure, IF $a+sqrtb$, where $a$ is a rational and $b$ is a nonsquare rational, is a root of a rational polynomial, you do get that $a-sqrtb$ 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.
– Batominovski
Jul 20 at 15:48
 |Â
show 1 more comment
up vote
1
down vote
up vote
1
down vote
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
$$beginalign a^2&+b^2+c^2+d^2+e^2
\&=(a+b+c+d+e)^2-2(ab+ac+ad+ae+bc+bd+be+cd+ce+de)
\&=0-2cdot 0=0,.endalign$$
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.
answered Jul 20 at 12:18
Batominovski
23.2k22777
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
$$beginalign a^2&+b^2+c^2+d^2+e^2
\&=(a+b+c+d+e)^2-2(ab+ac+ad+ae+bc+bd+be+cd+ce+de)
\&=0-2cdot 0=0,.endalign$$
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.
answered Jul 20 at 12:18
Batominovski
23.2k22777
edited Jul 20 at 12:24
answered Jul 20 at 12:18
Batominovski
23.2k22777
answered Jul 20 at 12:18
Batominovski
23.2k22777
answered Jul 20 at 12:18
Batominovski
23.2k22777
23.2k22777
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
– cmi
Jul 20 at 12:46
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?
– Batominovski
Jul 20 at 12:50
I am not asking about irrational roots. I am giving an argument. Can you please read my answer one more time?@Batominovski
– cmi
Jul 20 at 12:52
I replied. What does it mean for irrational numbers to be conjugates? This is not a quadratic polynomial. I think you are very confused.
– Batominovski
Jul 20 at 12:53
1
Sure, IF $a+sqrtb$, where $a$ is a rational and $b$ is a nonsquare rational, is a root of a rational polynomial, you do get that $a-sqrtb$ 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.
– Batominovski
Jul 20 at 15:48
 |Â
show 1 more comment
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
– cmi
Jul 20 at 12:46
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?
– Batominovski
Jul 20 at 12:50
I am not asking about irrational roots. I am giving an argument. Can you please read my answer one more time?@Batominovski
– cmi
Jul 20 at 12:52
I replied. What does it mean for irrational numbers to be conjugates? This is not a quadratic polynomial. I think you are very confused.
– Batominovski
Jul 20 at 12:53
1
Sure, IF $a+sqrtb$, where $a$ is a rational and $b$ is a nonsquare rational, is a root of a rational polynomial, you do get that $a-sqrtb$ 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.
– Batominovski
Jul 20 at 15:48
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
– cmi
Jul 20 at 12:46
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
– cmi
Jul 20 at 12:46
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?
– Batominovski
Jul 20 at 12:50
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?
– Batominovski
Jul 20 at 12:50
I am not asking about irrational roots. I am giving an argument. Can you please read my answer one more time?@Batominovski
– cmi
Jul 20 at 12:52
I am not asking about irrational roots. I am giving an argument. Can you please read my answer one more time?@Batominovski
– cmi
Jul 20 at 12:52
I replied. What does it mean for irrational numbers to be conjugates? This is not a quadratic polynomial. I think you are very confused.
– Batominovski
Jul 20 at 12:53
I replied. What does it mean for irrational numbers to be conjugates? This is not a quadratic polynomial. I think you are very confused.
– Batominovski
Jul 20 at 12:53
1
1
Sure, IF $a+sqrtb$, where $a$ is a rational and $b$ is a nonsquare rational, is a root of a rational polynomial, you do get that $a-sqrtb$ 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.
– Batominovski
Jul 20 at 15:48
Sure, IF $a+sqrtb$, where $a$ is a rational and $b$ is a nonsquare rational, is a root of a rational polynomial, you do get that $a-sqrtb$ 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.
– Batominovski
Jul 20 at 15:48
 |Â
show 1 more comment
up vote
1
down vote
Just adding to RobertZ's precise answer:
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.
answered Jul 20 at 12:36
Abcd
2,3761624
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.?
– cmi
Jul 20 at 12:45
@cmi Irrational roots need come in pairs
– Abcd
Jul 20 at 12:55
@cmi check this: math.stackexchange.com/questions/2608475/…
– Abcd
Jul 20 at 12:55
add a comment |Â
up vote
1
down vote
Just adding to RobertZ's precise answer:
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.
answered Jul 20 at 12:36
Abcd
2,3761624
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.?
– cmi
Jul 20 at 12:45
@cmi Irrational roots need come in pairs
– Abcd
Jul 20 at 12:55
@cmi check this: math.stackexchange.com/questions/2608475/…
– Abcd
Jul 20 at 12:55
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Just adding to RobertZ's precise answer:
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.
answered Jul 20 at 12:36
Abcd
2,3761624
Just adding to RobertZ's precise answer:
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.
answered Jul 20 at 12:36
Abcd
2,3761624
answered Jul 20 at 12:36
Abcd
2,3761624
answered Jul 20 at 12:36
Abcd
2,3761624
answered Jul 20 at 12:36
Abcd
2,3761624
2,3761624
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.?
– cmi
Jul 20 at 12:45
@cmi Irrational roots need come in pairs
– Abcd
Jul 20 at 12:55
@cmi check this: math.stackexchange.com/questions/2608475/…
– Abcd
Jul 20 at 12:55
add a comment |Â
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.?
– cmi
Jul 20 at 12:45
@cmi Irrational roots need come in pairs
– Abcd
Jul 20 at 12:55
@cmi check this: math.stackexchange.com/questions/2608475/…
– Abcd
Jul 20 at 12:55
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.?
– cmi
Jul 20 at 12:45
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.?
– cmi
Jul 20 at 12:45
@cmi Irrational roots need come in pairs
– Abcd
Jul 20 at 12:55
@cmi Irrational roots need come in pairs
– Abcd
Jul 20 at 12:55
@cmi check this: math.stackexchange.com/questions/2608475/…
– Abcd
Jul 20 at 12:55
@cmi check this: math.stackexchange.com/questions/2608475/…
– Abcd
Jul 20 at 12:55
add a comment |Â
up vote
1
down vote
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
$$ left( X^5-5X+2 , 5X^4 -5 , 4X-2 , frac7516 right). $$
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.
answered Jul 20 at 13:00
Chappers
55k74190
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.
– cmi
Jul 20 at 15:31
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
– cmi
Jul 20 at 15:31
add a comment |Â
up vote
1
down vote
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
$$ left( X^5-5X+2 , 5X^4 -5 , 4X-2 , frac7516 right). $$
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.
answered Jul 20 at 13:00
Chappers
55k74190
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.
– cmi
Jul 20 at 15:31
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
– cmi
Jul 20 at 15:31
add a comment |Â
up vote
1
down vote
up vote
1
down vote
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
$$ left( X^5-5X+2 , 5X^4 -5 , 4X-2 , frac7516 right). $$
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.
answered Jul 20 at 13:00
Chappers
55k74190
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
$$ left( X^5-5X+2 , 5X^4 -5 , 4X-2 , frac7516 right). $$
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.
answered Jul 20 at 13:00
Chappers
55k74190
answered Jul 20 at 13:00
Chappers
55k74190
answered Jul 20 at 13:00
Chappers
55k74190
answered Jul 20 at 13:00
Chappers
55k74190
55k74190
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.
– cmi
Jul 20 at 15:31
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
– cmi
Jul 20 at 15:31
add a comment |Â
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.
– cmi
Jul 20 at 15:31
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
– cmi
Jul 20 at 15:31
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.
– cmi
Jul 20 at 15:31
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.
– cmi
Jul 20 at 15:31
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
– cmi
Jul 20 at 15:31
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
– cmi
Jul 20 at 15:31
add a comment |Â
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1
So, do you know Rolle's Theorem? Or do you know Descartes' Rule of Sign?
– GEdgar
Jul 20 at 12:06
There is another real root in $(-2,-1)$.
– Batominovski
Jul 20 at 12:10