Determine the number of real solutions of an equation
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I need to calculate the number of real solutions in a polynomial equation. From the Fundamental Theorem of Algebra, I know that an equation of degree n has exactly n roots (but, here the complex ones are also included). And I just need to determine how many real ones.
For example, I am asked to the following equation:
$3x^5+15x-8=0$
From the Fundamental Algebra Theorem, it is known to have 5 roots (real and complex). Next, I have made a sketch of the function and it comes out that it only cuts the X-axis once between zero and one. Then, this equation would have a real solution and 4 complex solutions. But is there any way to solve it analytically without having to graph the function?
algebra-precalculus
asked Dec 25 '18 at 8:56
aprendiendo-a-programaraprendiendo-a-programar
83
$endgroup$
add a comment |
$begingroup$
I need to calculate the number of real solutions in a polynomial equation. From the Fundamental Theorem of Algebra, I know that an equation of degree n has exactly n roots (but, here the complex ones are also included). And I just need to determine how many real ones.
For example, I am asked to the following equation:
$3x^5+15x-8=0$
From the Fundamental Algebra Theorem, it is known to have 5 roots (real and complex). Next, I have made a sketch of the function and it comes out that it only cuts the X-axis once between zero and one. Then, this equation would have a real solution and 4 complex solutions. But is there any way to solve it analytically without having to graph the function?
algebra-precalculus
asked Dec 25 '18 at 8:56
aprendiendo-a-programaraprendiendo-a-programar
83
$endgroup$
5
$begingroup$
Hint: will have at least one real root because of odd degree of the polynomial. Then consider the derivative $15(x^4+1) >0$, so the function is increasing.
$endgroup$
– Anurag A
Dec 25 '18 at 8:58
1
$begingroup$
According to en.wikipedia.org/wiki/Descartes%27_rule_of_signs there's exactly one and that zero is positive.
$endgroup$
– Michael Hoppe
Dec 25 '18 at 19:00
add a comment |
$begingroup$
I need to calculate the number of real solutions in a polynomial equation. From the Fundamental Theorem of Algebra, I know that an equation of degree n has exactly n roots (but, here the complex ones are also included). And I just need to determine how many real ones.
For example, I am asked to the following equation:
$3x^5+15x-8=0$
From the Fundamental Algebra Theorem, it is known to have 5 roots (real and complex). Next, I have made a sketch of the function and it comes out that it only cuts the X-axis once between zero and one. Then, this equation would have a real solution and 4 complex solutions. But is there any way to solve it analytically without having to graph the function?
algebra-precalculus
asked Dec 25 '18 at 8:56
aprendiendo-a-programaraprendiendo-a-programar
83
$endgroup$
I need to calculate the number of real solutions in a polynomial equation. From the Fundamental Theorem of Algebra, I know that an equation of degree n has exactly n roots (but, here the complex ones are also included). And I just need to determine how many real ones.
For example, I am asked to the following equation:
$3x^5+15x-8=0$
From the Fundamental Algebra Theorem, it is known to have 5 roots (real and complex). Next, I have made a sketch of the function and it comes out that it only cuts the X-axis once between zero and one. Then, this equation would have a real solution and 4 complex solutions. But is there any way to solve it analytically without having to graph the function?
algebra-precalculus
algebra-precalculus
asked Dec 25 '18 at 8:56
aprendiendo-a-programaraprendiendo-a-programar
83
asked Dec 25 '18 at 8:56
aprendiendo-a-programaraprendiendo-a-programar
83
asked Dec 25 '18 at 8:56
aprendiendo-a-programaraprendiendo-a-programar
83
asked Dec 25 '18 at 8:56
aprendiendo-a-programaraprendiendo-a-programar
83
asked Dec 25 '18 at 8:56
aprendiendo-a-programaraprendiendo-a-programar
83
83
5
$begingroup$
Hint: will have at least one real root because of odd degree of the polynomial. Then consider the derivative $15(x^4+1) >0$, so the function is increasing.
$endgroup$
– Anurag A
Dec 25 '18 at 8:58
1
$begingroup$
According to en.wikipedia.org/wiki/Descartes%27_rule_of_signs there's exactly one and that zero is positive.
$endgroup$
– Michael Hoppe
Dec 25 '18 at 19:00
add a comment |
5
$begingroup$
Hint: will have at least one real root because of odd degree of the polynomial. Then consider the derivative $15(x^4+1) >0$, so the function is increasing.
$endgroup$
– Anurag A
Dec 25 '18 at 8:58
1
$begingroup$
According to en.wikipedia.org/wiki/Descartes%27_rule_of_signs there's exactly one and that zero is positive.
$endgroup$
– Michael Hoppe
Dec 25 '18 at 19:00
5
5
$begingroup$
Hint: will have at least one real root because of odd degree of the polynomial. Then consider the derivative $15(x^4+1) >0$, so the function is increasing.
$endgroup$
– Anurag A
Dec 25 '18 at 8:58
$begingroup$
Hint: will have at least one real root because of odd degree of the polynomial. Then consider the derivative $15(x^4+1) >0$, so the function is increasing.
$endgroup$
– Anurag A
Dec 25 '18 at 8:58
1
1
$begingroup$
According to en.wikipedia.org/wiki/Descartes%27_rule_of_signs there's exactly one and that zero is positive.
$endgroup$
– Michael Hoppe
Dec 25 '18 at 19:00
$begingroup$
According to en.wikipedia.org/wiki/Descartes%27_rule_of_signs there's exactly one and that zero is positive.
$endgroup$
– Michael Hoppe
Dec 25 '18 at 19:00
add a comment |
1 Answer
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Let $P(x)=3x^5+15x-8$. Since $lim_{xtoinfty}P(x)=infty$ and $lim_{xto-infty}P(x)=-infty$, it follows from the intermediate value theorem that $P(x)$ has at least one real root. Now, imagine that it has two distinct roots $x_1$ and $x_2$. Then, by Rolle's theorem, there is some $y$ between $x_1$ and $x_2$ such that $P'(y)=0$. But $P'(y)=15y^4+15$, and therefore it can't be $0$.
answered Dec 25 '18 at 9:14
José Carlos SantosJosé Carlos Santos
178k24139251
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add a comment |
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$begingroup$
Let $P(x)=3x^5+15x-8$. Since $lim_{xtoinfty}P(x)=infty$ and $lim_{xto-infty}P(x)=-infty$, it follows from the intermediate value theorem that $P(x)$ has at least one real root. Now, imagine that it has two distinct roots $x_1$ and $x_2$. Then, by Rolle's theorem, there is some $y$ between $x_1$ and $x_2$ such that $P'(y)=0$. But $P'(y)=15y^4+15$, and therefore it can't be $0$.
answered Dec 25 '18 at 9:14
José Carlos SantosJosé Carlos Santos
178k24139251
$endgroup$
add a comment |
$begingroup$
Let $P(x)=3x^5+15x-8$. Since $lim_{xtoinfty}P(x)=infty$ and $lim_{xto-infty}P(x)=-infty$, it follows from the intermediate value theorem that $P(x)$ has at least one real root. Now, imagine that it has two distinct roots $x_1$ and $x_2$. Then, by Rolle's theorem, there is some $y$ between $x_1$ and $x_2$ such that $P'(y)=0$. But $P'(y)=15y^4+15$, and therefore it can't be $0$.
answered Dec 25 '18 at 9:14
José Carlos SantosJosé Carlos Santos
178k24139251
$endgroup$
add a comment |
$begingroup$
Let $P(x)=3x^5+15x-8$. Since $lim_{xtoinfty}P(x)=infty$ and $lim_{xto-infty}P(x)=-infty$, it follows from the intermediate value theorem that $P(x)$ has at least one real root. Now, imagine that it has two distinct roots $x_1$ and $x_2$. Then, by Rolle's theorem, there is some $y$ between $x_1$ and $x_2$ such that $P'(y)=0$. But $P'(y)=15y^4+15$, and therefore it can't be $0$.
answered Dec 25 '18 at 9:14
José Carlos SantosJosé Carlos Santos
178k24139251
$endgroup$
Let $P(x)=3x^5+15x-8$. Since $lim_{xtoinfty}P(x)=infty$ and $lim_{xto-infty}P(x)=-infty$, it follows from the intermediate value theorem that $P(x)$ has at least one real root. Now, imagine that it has two distinct roots $x_1$ and $x_2$. Then, by Rolle's theorem, there is some $y$ between $x_1$ and $x_2$ such that $P'(y)=0$. But $P'(y)=15y^4+15$, and therefore it can't be $0$.
answered Dec 25 '18 at 9:14
José Carlos SantosJosé Carlos Santos
178k24139251
answered Dec 25 '18 at 9:14
José Carlos SantosJosé Carlos Santos
178k24139251
answered Dec 25 '18 at 9:14
José Carlos SantosJosé Carlos Santos
178k24139251
answered Dec 25 '18 at 9:14
José Carlos SantosJosé Carlos Santos
178k24139251
178k24139251
add a comment |
add a comment |
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Hint: will have at least one real root because of odd degree of the polynomial. Then consider the derivative $15(x^4+1) >0$, so the function is increasing.
$endgroup$
– Anurag A
Dec 25 '18 at 8:58
1
$begingroup$
According to en.wikipedia.org/wiki/Descartes%27_rule_of_signs there's exactly one and that zero is positive.
$endgroup$
– Michael Hoppe
Dec 25 '18 at 19:00