I've solved an equation in two different ways, and I keep getting two different solutions, what's wrong?
Can you tell me where the mistake is? 
problem-solving
asked yesterday
Ashraf Benmebarek
435
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Can you tell me where the mistake is?
problem-solving
asked yesterday
Ashraf Benmebarek
435
add a comment |
Can you tell me where the mistake is?
problem-solving
asked yesterday
Ashraf Benmebarek
435
Can you tell me where the mistake is?
problem-solving
problem-solving
asked yesterday
Ashraf Benmebarek
435
asked yesterday
Ashraf Benmebarek
435
asked yesterday
Ashraf Benmebarek
435
asked yesterday
Ashraf Benmebarek
435
asked yesterday
Ashraf Benmebarek
435
435
add a comment |
add a comment |
3 Answers
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In the first method, you made the following mistake:
$$x^4cdot x^3 = x^7 color{red}{neq x^{12}}$$
The following statement is what you meant (not that it helps in any way here):
$$left(x^4right)^3 = x^{12}$$
Instead, you have $x^4cdot x^8 = x^{12}$, which yields
$$frac{6}{x^4} = frac{6x^8}{x^{12}} implies frac{6x^8-1536}{x^{12}} = 0; quad x^{12} neq 0$$
from which you obtain the desired result $x = pm 2$. Note that you missed this in your second way, hence the solution isn’t complete. Any equation in the form $x^2 = a$ has two solutions: $x = pmsqrt{a}$.
answered yesterday
KM101
5,3811423
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Your error is on the left, on the second line: $dfrac{6}{x^4} neq dfrac{6x^3}{x^{12}}$.
Correcting this, that second line should read $$dfrac{6x^8 - 1536}{x^{12}} = 0,$$ so $6x^8 = 1536$, and $x^8 = 256$, so $x = 2$ (or $x = -2$, which solution you missed in both attempts).
edited yesterday
Acccumulation
6,8062618
answered yesterday
user3482749
2,728414
add a comment |
$$frac{x^n}{x^m}=x^{n-m}impliesfrac{x^{12}}{x^4}=x^8$$
answered yesterday
DonAntonio
177k1492225
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
In the first method, you made the following mistake:
$$x^4cdot x^3 = x^7 color{red}{neq x^{12}}$$
The following statement is what you meant (not that it helps in any way here):
$$left(x^4right)^3 = x^{12}$$
Instead, you have $x^4cdot x^8 = x^{12}$, which yields
$$frac{6}{x^4} = frac{6x^8}{x^{12}} implies frac{6x^8-1536}{x^{12}} = 0; quad x^{12} neq 0$$
from which you obtain the desired result $x = pm 2$. Note that you missed this in your second way, hence the solution isn’t complete. Any equation in the form $x^2 = a$ has two solutions: $x = pmsqrt{a}$.
answered yesterday
KM101
5,3811423
add a comment |
In the first method, you made the following mistake:
$$x^4cdot x^3 = x^7 color{red}{neq x^{12}}$$
The following statement is what you meant (not that it helps in any way here):
$$left(x^4right)^3 = x^{12}$$
Instead, you have $x^4cdot x^8 = x^{12}$, which yields
$$frac{6}{x^4} = frac{6x^8}{x^{12}} implies frac{6x^8-1536}{x^{12}} = 0; quad x^{12} neq 0$$
from which you obtain the desired result $x = pm 2$. Note that you missed this in your second way, hence the solution isn’t complete. Any equation in the form $x^2 = a$ has two solutions: $x = pmsqrt{a}$.
answered yesterday
KM101
5,3811423
add a comment |
In the first method, you made the following mistake:
$$x^4cdot x^3 = x^7 color{red}{neq x^{12}}$$
The following statement is what you meant (not that it helps in any way here):
$$left(x^4right)^3 = x^{12}$$
Instead, you have $x^4cdot x^8 = x^{12}$, which yields
$$frac{6}{x^4} = frac{6x^8}{x^{12}} implies frac{6x^8-1536}{x^{12}} = 0; quad x^{12} neq 0$$
from which you obtain the desired result $x = pm 2$. Note that you missed this in your second way, hence the solution isn’t complete. Any equation in the form $x^2 = a$ has two solutions: $x = pmsqrt{a}$.
answered yesterday
KM101
5,3811423
In the first method, you made the following mistake:
$$x^4cdot x^3 = x^7 color{red}{neq x^{12}}$$
The following statement is what you meant (not that it helps in any way here):
$$left(x^4right)^3 = x^{12}$$
Instead, you have $x^4cdot x^8 = x^{12}$, which yields
$$frac{6}{x^4} = frac{6x^8}{x^{12}} implies frac{6x^8-1536}{x^{12}} = 0; quad x^{12} neq 0$$
from which you obtain the desired result $x = pm 2$. Note that you missed this in your second way, hence the solution isn’t complete. Any equation in the form $x^2 = a$ has two solutions: $x = pmsqrt{a}$.
answered yesterday
KM101
5,3811423
edited yesterday
answered yesterday
KM101
5,3811423
answered yesterday
KM101
5,3811423
answered yesterday
KM101
5,3811423
5,3811423
add a comment |
add a comment |
Your error is on the left, on the second line: $dfrac{6}{x^4} neq dfrac{6x^3}{x^{12}}$.
Correcting this, that second line should read $$dfrac{6x^8 - 1536}{x^{12}} = 0,$$ so $6x^8 = 1536$, and $x^8 = 256$, so $x = 2$ (or $x = -2$, which solution you missed in both attempts).
edited yesterday
Acccumulation
6,8062618
answered yesterday
user3482749
2,728414
add a comment |
Your error is on the left, on the second line: $dfrac{6}{x^4} neq dfrac{6x^3}{x^{12}}$.
Correcting this, that second line should read $$dfrac{6x^8 - 1536}{x^{12}} = 0,$$ so $6x^8 = 1536$, and $x^8 = 256$, so $x = 2$ (or $x = -2$, which solution you missed in both attempts).
edited yesterday
Acccumulation
6,8062618
answered yesterday
user3482749
2,728414
add a comment |
Your error is on the left, on the second line: $dfrac{6}{x^4} neq dfrac{6x^3}{x^{12}}$.
Correcting this, that second line should read $$dfrac{6x^8 - 1536}{x^{12}} = 0,$$ so $6x^8 = 1536$, and $x^8 = 256$, so $x = 2$ (or $x = -2$, which solution you missed in both attempts).
edited yesterday
Acccumulation
6,8062618
answered yesterday
user3482749
2,728414
Your error is on the left, on the second line: $dfrac{6}{x^4} neq dfrac{6x^3}{x^{12}}$.
Correcting this, that second line should read $$dfrac{6x^8 - 1536}{x^{12}} = 0,$$ so $6x^8 = 1536$, and $x^8 = 256$, so $x = 2$ (or $x = -2$, which solution you missed in both attempts).
edited yesterday
Acccumulation
6,8062618
answered yesterday
user3482749
2,728414
edited yesterday
Acccumulation
6,8062618
edited yesterday
Acccumulation
6,8062618
edited yesterday
Acccumulation
6,8062618
6,8062618
answered yesterday
user3482749
2,728414
answered yesterday
user3482749
2,728414
answered yesterday
user3482749
2,728414
2,728414
add a comment |
add a comment |
$$frac{x^n}{x^m}=x^{n-m}impliesfrac{x^{12}}{x^4}=x^8$$
answered yesterday
DonAntonio
177k1492225
add a comment |
$$frac{x^n}{x^m}=x^{n-m}impliesfrac{x^{12}}{x^4}=x^8$$
answered yesterday
DonAntonio
177k1492225
add a comment |
$$frac{x^n}{x^m}=x^{n-m}impliesfrac{x^{12}}{x^4}=x^8$$
answered yesterday
DonAntonio
177k1492225
$$frac{x^n}{x^m}=x^{n-m}impliesfrac{x^{12}}{x^4}=x^8$$
answered yesterday
DonAntonio
177k1492225
answered yesterday
DonAntonio
177k1492225
answered yesterday
DonAntonio
177k1492225
answered yesterday
DonAntonio
177k1492225
177k1492225
add a comment |
add a comment |
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