How to solve $x^6- 4x^3+4=0$?
$begingroup$
I'm working through an exercise with the question:
Use substitution to solve for $x$ in the following equation
$x^6 - 4x^3 + 4 = 0$
The solution given is $sqrt[3]2$ however there is no working in between shown and I'm at a loss for how to arrive at this.
I started trying to factor our the polynomials $x^6$ and $4x^3$ but that didn't really get me anywhere:
$x^3(x^3 - 4) = -4$ is as far as I got.
How can I arrive at $sqrt[3]2$ with each step shown in between?
algebra-precalculus polynomials roots
edited Dec 11 '18 at 17:19
Brahadeesh
6,46942363
asked Dec 11 '18 at 17:04
Doug FirDoug Fir
3418
$endgroup$
add a comment |
$begingroup$
I'm working through an exercise with the question:
Use substitution to solve for $x$ in the following equation
$x^6 - 4x^3 + 4 = 0$
The solution given is $sqrt[3]2$ however there is no working in between shown and I'm at a loss for how to arrive at this.
I started trying to factor our the polynomials $x^6$ and $4x^3$ but that didn't really get me anywhere:
$x^3(x^3 - 4) = -4$ is as far as I got.
How can I arrive at $sqrt[3]2$ with each step shown in between?
algebra-precalculus polynomials roots
edited Dec 11 '18 at 17:19
Brahadeesh
6,46942363
asked Dec 11 '18 at 17:04
Doug FirDoug Fir
3418
$endgroup$
5
$begingroup$
Let $u=x^3$. Then your equation will turn into a quadratic equation.
$endgroup$
– user9077
Dec 11 '18 at 17:07
add a comment |
$begingroup$
I'm working through an exercise with the question:
Use substitution to solve for $x$ in the following equation
$x^6 - 4x^3 + 4 = 0$
The solution given is $sqrt[3]2$ however there is no working in between shown and I'm at a loss for how to arrive at this.
I started trying to factor our the polynomials $x^6$ and $4x^3$ but that didn't really get me anywhere:
$x^3(x^3 - 4) = -4$ is as far as I got.
How can I arrive at $sqrt[3]2$ with each step shown in between?
algebra-precalculus polynomials roots
edited Dec 11 '18 at 17:19
Brahadeesh
6,46942363
asked Dec 11 '18 at 17:04
Doug FirDoug Fir
3418
$endgroup$
I'm working through an exercise with the question:
Use substitution to solve for $x$ in the following equation
$x^6 - 4x^3 + 4 = 0$
The solution given is $sqrt[3]2$ however there is no working in between shown and I'm at a loss for how to arrive at this.
I started trying to factor our the polynomials $x^6$ and $4x^3$ but that didn't really get me anywhere:
$x^3(x^3 - 4) = -4$ is as far as I got.
How can I arrive at $sqrt[3]2$ with each step shown in between?
algebra-precalculus polynomials roots
algebra-precalculus polynomials roots
edited Dec 11 '18 at 17:19
Brahadeesh
6,46942363
asked Dec 11 '18 at 17:04
Doug FirDoug Fir
3418
edited Dec 11 '18 at 17:19
Brahadeesh
6,46942363
asked Dec 11 '18 at 17:04
Doug FirDoug Fir
3418
edited Dec 11 '18 at 17:19
Brahadeesh
6,46942363
edited Dec 11 '18 at 17:19
Brahadeesh
6,46942363
edited Dec 11 '18 at 17:19
Brahadeesh
6,46942363
6,46942363
asked Dec 11 '18 at 17:04
Doug FirDoug Fir
3418
asked Dec 11 '18 at 17:04
Doug FirDoug Fir
3418
asked Dec 11 '18 at 17:04
Doug FirDoug Fir
3418
3418
5
$begingroup$
Let $u=x^3$. Then your equation will turn into a quadratic equation.
$endgroup$
– user9077
Dec 11 '18 at 17:07
add a comment |
5
$begingroup$
Let $u=x^3$. Then your equation will turn into a quadratic equation.
$endgroup$
– user9077
Dec 11 '18 at 17:07
5
5
$begingroup$
Let $u=x^3$. Then your equation will turn into a quadratic equation.
$endgroup$
– user9077
Dec 11 '18 at 17:07
$begingroup$
Let $u=x^3$. Then your equation will turn into a quadratic equation.
$endgroup$
– user9077
Dec 11 '18 at 17:07
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: put $y = x^3$, and you get a quadratic equation.
answered Dec 11 '18 at 17:07
DeepSeaDeepSea
71.3k54487
$endgroup$
add a comment |
$begingroup$
Use that $x^6-4x^3+4=(x^3-2)^2$
answered Dec 11 '18 at 17:12
VasyaVasya
3,3771516
$endgroup$
add a comment |
$begingroup$
A simple substitution gets the job done. Let $t = x^3$.
$$x^3(x^3-4) = -4 iff t(t-4) = -4 iff t = 2$$
$$t = 2 iff x^3 = 2 iff x = sqrt[3]{2}$$
As a more general tip, whenever you have “quadratic-like” equations, make a substitution to reach a normal quadratic.
$$ax^{2n}+bx^n+c = 0$$
Here, letting $t = x^n$ reduces the problem to solving for $t$ in $at^2+bt+c = 0$.
answered Dec 11 '18 at 17:15
KM101KM101
6,0251525
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: put $y = x^3$, and you get a quadratic equation.
answered Dec 11 '18 at 17:07
DeepSeaDeepSea
71.3k54487
$endgroup$
add a comment |
$begingroup$
Hint: put $y = x^3$, and you get a quadratic equation.
answered Dec 11 '18 at 17:07
DeepSeaDeepSea
71.3k54487
$endgroup$
add a comment |
$begingroup$
Hint: put $y = x^3$, and you get a quadratic equation.
answered Dec 11 '18 at 17:07
DeepSeaDeepSea
71.3k54487
$endgroup$
Hint: put $y = x^3$, and you get a quadratic equation.
answered Dec 11 '18 at 17:07
DeepSeaDeepSea
71.3k54487
answered Dec 11 '18 at 17:07
DeepSeaDeepSea
71.3k54487
answered Dec 11 '18 at 17:07
DeepSeaDeepSea
71.3k54487
answered Dec 11 '18 at 17:07
DeepSeaDeepSea
71.3k54487
71.3k54487
add a comment |
add a comment |
$begingroup$
Use that $x^6-4x^3+4=(x^3-2)^2$
answered Dec 11 '18 at 17:12
VasyaVasya
3,3771516
$endgroup$
add a comment |
$begingroup$
Use that $x^6-4x^3+4=(x^3-2)^2$
answered Dec 11 '18 at 17:12
VasyaVasya
3,3771516
$endgroup$
add a comment |
$begingroup$
Use that $x^6-4x^3+4=(x^3-2)^2$
answered Dec 11 '18 at 17:12
VasyaVasya
3,3771516
$endgroup$
Use that $x^6-4x^3+4=(x^3-2)^2$
answered Dec 11 '18 at 17:12
VasyaVasya
3,3771516
answered Dec 11 '18 at 17:12
VasyaVasya
3,3771516
answered Dec 11 '18 at 17:12
VasyaVasya
3,3771516
answered Dec 11 '18 at 17:12
VasyaVasya
3,3771516
3,3771516
add a comment |
add a comment |
$begingroup$
A simple substitution gets the job done. Let $t = x^3$.
$$x^3(x^3-4) = -4 iff t(t-4) = -4 iff t = 2$$
$$t = 2 iff x^3 = 2 iff x = sqrt[3]{2}$$
As a more general tip, whenever you have “quadratic-like” equations, make a substitution to reach a normal quadratic.
$$ax^{2n}+bx^n+c = 0$$
Here, letting $t = x^n$ reduces the problem to solving for $t$ in $at^2+bt+c = 0$.
answered Dec 11 '18 at 17:15
KM101KM101
6,0251525
$endgroup$
add a comment |
$begingroup$
A simple substitution gets the job done. Let $t = x^3$.
$$x^3(x^3-4) = -4 iff t(t-4) = -4 iff t = 2$$
$$t = 2 iff x^3 = 2 iff x = sqrt[3]{2}$$
As a more general tip, whenever you have “quadratic-like” equations, make a substitution to reach a normal quadratic.
$$ax^{2n}+bx^n+c = 0$$
Here, letting $t = x^n$ reduces the problem to solving for $t$ in $at^2+bt+c = 0$.
answered Dec 11 '18 at 17:15
KM101KM101
6,0251525
$endgroup$
add a comment |
$begingroup$
A simple substitution gets the job done. Let $t = x^3$.
$$x^3(x^3-4) = -4 iff t(t-4) = -4 iff t = 2$$
$$t = 2 iff x^3 = 2 iff x = sqrt[3]{2}$$
As a more general tip, whenever you have “quadratic-like” equations, make a substitution to reach a normal quadratic.
$$ax^{2n}+bx^n+c = 0$$
Here, letting $t = x^n$ reduces the problem to solving for $t$ in $at^2+bt+c = 0$.
answered Dec 11 '18 at 17:15
KM101KM101
6,0251525
$endgroup$
A simple substitution gets the job done. Let $t = x^3$.
$$x^3(x^3-4) = -4 iff t(t-4) = -4 iff t = 2$$
$$t = 2 iff x^3 = 2 iff x = sqrt[3]{2}$$
As a more general tip, whenever you have “quadratic-like” equations, make a substitution to reach a normal quadratic.
$$ax^{2n}+bx^n+c = 0$$
Here, letting $t = x^n$ reduces the problem to solving for $t$ in $at^2+bt+c = 0$.
answered Dec 11 '18 at 17:15
KM101KM101
6,0251525
edited Dec 11 '18 at 17:21
answered Dec 11 '18 at 17:15
KM101KM101
6,0251525
answered Dec 11 '18 at 17:15
KM101KM101
6,0251525
answered Dec 11 '18 at 17:15
KM101KM101
6,0251525
6,0251525
add a comment |
add a comment |
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5
$begingroup$
Let $u=x^3$. Then your equation will turn into a quadratic equation.
$endgroup$
– user9077
Dec 11 '18 at 17:07