I've solved an equation in two different ways, and I keep getting two different solutions, what's wrong?
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Can you tell me where the mistake is? 
problem-solving
asked Jan 4 at 20:18
Ashraf BenmebarekAshraf Benmebarek
465
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$begingroup$
Can you tell me where the mistake is?
problem-solving
asked Jan 4 at 20:18
Ashraf BenmebarekAshraf Benmebarek
465
$endgroup$
add a comment |
$begingroup$
Can you tell me where the mistake is?
problem-solving
asked Jan 4 at 20:18
Ashraf BenmebarekAshraf Benmebarek
465
$endgroup$
Can you tell me where the mistake is?
problem-solving
problem-solving
asked Jan 4 at 20:18
Ashraf BenmebarekAshraf Benmebarek
465
asked Jan 4 at 20:18
Ashraf BenmebarekAshraf Benmebarek
465
asked Jan 4 at 20:18
Ashraf BenmebarekAshraf Benmebarek
465
asked Jan 4 at 20:18
Ashraf BenmebarekAshraf Benmebarek
465
asked Jan 4 at 20:18
Ashraf BenmebarekAshraf Benmebarek
465
465
add a comment |
add a comment |
3 Answers
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$begingroup$
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 Jan 4 at 20:24
KM101KM101
6,0701525
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$begingroup$
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 Jan 4 at 20:23
Acccumulation
7,0852619
answered Jan 4 at 20:22
user3482749user3482749
4,296919
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$$frac{x^n}{x^m}=x^{n-m}impliesfrac{x^{12}}{x^4}=x^8$$
answered Jan 4 at 20:24
DonAntonioDonAntonio
179k1494232
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Jan 4 at 20:24
KM101KM101
6,0701525
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 20:24
KM101KM101
6,0701525
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 20:24
KM101KM101
6,0701525
$endgroup$
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 Jan 4 at 20:24
KM101KM101
6,0701525
edited Jan 4 at 20:29
answered Jan 4 at 20:24
KM101KM101
6,0701525
answered Jan 4 at 20:24
KM101KM101
6,0701525
answered Jan 4 at 20:24
KM101KM101
6,0701525
6,0701525
add a comment |
add a comment |
$begingroup$
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 Jan 4 at 20:23
Acccumulation
7,0852619
answered Jan 4 at 20:22
user3482749user3482749
4,296919
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 20:23
Acccumulation
7,0852619
answered Jan 4 at 20:22
user3482749user3482749
4,296919
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 20:23
Acccumulation
7,0852619
answered Jan 4 at 20:22
user3482749user3482749
4,296919
$endgroup$
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 Jan 4 at 20:23
Acccumulation
7,0852619
answered Jan 4 at 20:22
user3482749user3482749
4,296919
edited Jan 4 at 20:23
Acccumulation
7,0852619
edited Jan 4 at 20:23
Acccumulation
7,0852619
edited Jan 4 at 20:23
Acccumulation
7,0852619
7,0852619
answered Jan 4 at 20:22
user3482749user3482749
4,296919
answered Jan 4 at 20:22
user3482749user3482749
4,296919
answered Jan 4 at 20:22
user3482749user3482749
4,296919
4,296919
add a comment |
add a comment |
$begingroup$
$$frac{x^n}{x^m}=x^{n-m}impliesfrac{x^{12}}{x^4}=x^8$$
answered Jan 4 at 20:24
DonAntonioDonAntonio
179k1494232
$endgroup$
add a comment |
$begingroup$
$$frac{x^n}{x^m}=x^{n-m}impliesfrac{x^{12}}{x^4}=x^8$$
answered Jan 4 at 20:24
DonAntonioDonAntonio
179k1494232
$endgroup$
add a comment |
$begingroup$
$$frac{x^n}{x^m}=x^{n-m}impliesfrac{x^{12}}{x^4}=x^8$$
answered Jan 4 at 20:24
DonAntonioDonAntonio
179k1494232
$endgroup$
$$frac{x^n}{x^m}=x^{n-m}impliesfrac{x^{12}}{x^4}=x^8$$
answered Jan 4 at 20:24
DonAntonioDonAntonio
179k1494232
answered Jan 4 at 20:24
DonAntonioDonAntonio
179k1494232
answered Jan 4 at 20:24
DonAntonioDonAntonio
179k1494232
answered Jan 4 at 20:24
DonAntonioDonAntonio
179k1494232
179k1494232
add a comment |
add a comment |
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