Solve $(x^2-5x+5)^x^2-7x+12= 1$.
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What are all the possible values of $x$ if $(x^2-5x+5)^x^2-7x+12= 1$. I have already found three answers: 4, 3, and 1, however, apparently there are more possibilities. I don't know how to figure this out so it would be extremely appreciated if someone found the other possibilities and showed me how to do it.
quadratics problem-solving integers
add a comment |Â
up vote
3
down vote
favorite
What are all the possible values of $x$ if $(x^2-5x+5)^x^2-7x+12= 1$. I have already found three answers: 4, 3, and 1, however, apparently there are more possibilities. I don't know how to figure this out so it would be extremely appreciated if someone found the other possibilities and showed me how to do it.
quadratics problem-solving integers
1
Please check the equation. The parentheses don't balance. (While you're editing, put the equation into the title of your question.)
â Blue
Aug 31 at 1:18
1
I have adjusted the problem's typesetting to be consistent with the $4,3,1$ solutions given.
â vadim123
Aug 31 at 1:29
$x^2-7x+12 = (x-3)(x-4)$
â amsmath
Aug 31 at 1:31
For positive $x^2-5x+5$ your solutions are the only ones. The question is whether negative $x^2-5x+5$ is allowed.
â amsmath
Aug 31 at 1:38
No, wait. For $x^2-5x+5 < 0$ for $x=3$.
â amsmath
Aug 31 at 1:52
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
What are all the possible values of $x$ if $(x^2-5x+5)^x^2-7x+12= 1$. I have already found three answers: 4, 3, and 1, however, apparently there are more possibilities. I don't know how to figure this out so it would be extremely appreciated if someone found the other possibilities and showed me how to do it.
quadratics problem-solving integers
What are all the possible values of $x$ if $(x^2-5x+5)^x^2-7x+12= 1$. I have already found three answers: 4, 3, and 1, however, apparently there are more possibilities. I don't know how to figure this out so it would be extremely appreciated if someone found the other possibilities and showed me how to do it.
quadratics problem-solving integers
quadratics problem-solving integers
edited Aug 31 at 8:10
tarit goswami
1,158219
1,158219
asked Aug 31 at 1:11
Julie
364
364
1
Please check the equation. The parentheses don't balance. (While you're editing, put the equation into the title of your question.)
â Blue
Aug 31 at 1:18
1
I have adjusted the problem's typesetting to be consistent with the $4,3,1$ solutions given.
â vadim123
Aug 31 at 1:29
$x^2-7x+12 = (x-3)(x-4)$
â amsmath
Aug 31 at 1:31
For positive $x^2-5x+5$ your solutions are the only ones. The question is whether negative $x^2-5x+5$ is allowed.
â amsmath
Aug 31 at 1:38
No, wait. For $x^2-5x+5 < 0$ for $x=3$.
â amsmath
Aug 31 at 1:52
add a comment |Â
1
Please check the equation. The parentheses don't balance. (While you're editing, put the equation into the title of your question.)
â Blue
Aug 31 at 1:18
1
I have adjusted the problem's typesetting to be consistent with the $4,3,1$ solutions given.
â vadim123
Aug 31 at 1:29
$x^2-7x+12 = (x-3)(x-4)$
â amsmath
Aug 31 at 1:31
For positive $x^2-5x+5$ your solutions are the only ones. The question is whether negative $x^2-5x+5$ is allowed.
â amsmath
Aug 31 at 1:38
No, wait. For $x^2-5x+5 < 0$ for $x=3$.
â amsmath
Aug 31 at 1:52
1
1
Please check the equation. The parentheses don't balance. (While you're editing, put the equation into the title of your question.)
â Blue
Aug 31 at 1:18
Please check the equation. The parentheses don't balance. (While you're editing, put the equation into the title of your question.)
â Blue
Aug 31 at 1:18
1
1
I have adjusted the problem's typesetting to be consistent with the $4,3,1$ solutions given.
â vadim123
Aug 31 at 1:29
I have adjusted the problem's typesetting to be consistent with the $4,3,1$ solutions given.
â vadim123
Aug 31 at 1:29
$x^2-7x+12 = (x-3)(x-4)$
â amsmath
Aug 31 at 1:31
$x^2-7x+12 = (x-3)(x-4)$
â amsmath
Aug 31 at 1:31
For positive $x^2-5x+5$ your solutions are the only ones. The question is whether negative $x^2-5x+5$ is allowed.
â amsmath
Aug 31 at 1:38
For positive $x^2-5x+5$ your solutions are the only ones. The question is whether negative $x^2-5x+5$ is allowed.
â amsmath
Aug 31 at 1:38
No, wait. For $x^2-5x+5 < 0$ for $x=3$.
â amsmath
Aug 31 at 1:52
No, wait. For $x^2-5x+5 < 0$ for $x=3$.
â amsmath
Aug 31 at 1:52
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
Hint:
$(-1)^k=1$ (if $k$ is an even integer).
In other words, you forgot the case that $x^2-5x+5=-1.$
Solving this, we get $x^2-5x+6=0rightarrow x=2,3.$
We check that the exponent is even.
If $x=3,$ then $(x-3)(x-4)=0.$
If $x=2,$ then $(-1)(-2)=2.$
Either way, it is even, so both solutions are valid.
All solutions are $ 1,2,3,4$.
So how would you use this to get the other possibilities? Sorry, I'm still quite confused.
â Julie
Aug 31 at 1:33
Solve for $x^2-5x+5=-1$ and check the exponent is even.
â Jason Kim
Aug 31 at 1:34
In particular ~ there are three possibilities: The exponent is $0,$ the base is $1/-1.$ $-1$ is the case that was not included in the solution.
â Jason Kim
Aug 31 at 1:37
1
@Julie You should actually answer this question yourself. Do you just have this equation or any more information? For example, is the question to find all integer values of $x$ solving the equation?
â amsmath
Aug 31 at 1:59
1
@JasonKim Recently in an entrance exam it was asked to explain why these are all solutions and there is no other(in first step asked to find the solutions). I have found all, but was not able to explain properly(I have explained by taking $log$ both side) why these are all solutions and there are no other. Can you explain :How we are sure there are no other solutions
?
â tarit goswami
Aug 31 at 4:57
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Hint:
$(-1)^k=1$ (if $k$ is an even integer).
In other words, you forgot the case that $x^2-5x+5=-1.$
Solving this, we get $x^2-5x+6=0rightarrow x=2,3.$
We check that the exponent is even.
If $x=3,$ then $(x-3)(x-4)=0.$
If $x=2,$ then $(-1)(-2)=2.$
Either way, it is even, so both solutions are valid.
All solutions are $ 1,2,3,4$.
So how would you use this to get the other possibilities? Sorry, I'm still quite confused.
â Julie
Aug 31 at 1:33
Solve for $x^2-5x+5=-1$ and check the exponent is even.
â Jason Kim
Aug 31 at 1:34
In particular ~ there are three possibilities: The exponent is $0,$ the base is $1/-1.$ $-1$ is the case that was not included in the solution.
â Jason Kim
Aug 31 at 1:37
1
@Julie You should actually answer this question yourself. Do you just have this equation or any more information? For example, is the question to find all integer values of $x$ solving the equation?
â amsmath
Aug 31 at 1:59
1
@JasonKim Recently in an entrance exam it was asked to explain why these are all solutions and there is no other(in first step asked to find the solutions). I have found all, but was not able to explain properly(I have explained by taking $log$ both side) why these are all solutions and there are no other. Can you explain :How we are sure there are no other solutions
?
â tarit goswami
Aug 31 at 4:57
 |Â
show 3 more comments
up vote
5
down vote
accepted
Hint:
$(-1)^k=1$ (if $k$ is an even integer).
In other words, you forgot the case that $x^2-5x+5=-1.$
Solving this, we get $x^2-5x+6=0rightarrow x=2,3.$
We check that the exponent is even.
If $x=3,$ then $(x-3)(x-4)=0.$
If $x=2,$ then $(-1)(-2)=2.$
Either way, it is even, so both solutions are valid.
All solutions are $ 1,2,3,4$.
So how would you use this to get the other possibilities? Sorry, I'm still quite confused.
â Julie
Aug 31 at 1:33
Solve for $x^2-5x+5=-1$ and check the exponent is even.
â Jason Kim
Aug 31 at 1:34
In particular ~ there are three possibilities: The exponent is $0,$ the base is $1/-1.$ $-1$ is the case that was not included in the solution.
â Jason Kim
Aug 31 at 1:37
1
@Julie You should actually answer this question yourself. Do you just have this equation or any more information? For example, is the question to find all integer values of $x$ solving the equation?
â amsmath
Aug 31 at 1:59
1
@JasonKim Recently in an entrance exam it was asked to explain why these are all solutions and there is no other(in first step asked to find the solutions). I have found all, but was not able to explain properly(I have explained by taking $log$ both side) why these are all solutions and there are no other. Can you explain :How we are sure there are no other solutions
?
â tarit goswami
Aug 31 at 4:57
 |Â
show 3 more comments
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Hint:
$(-1)^k=1$ (if $k$ is an even integer).
In other words, you forgot the case that $x^2-5x+5=-1.$
Solving this, we get $x^2-5x+6=0rightarrow x=2,3.$
We check that the exponent is even.
If $x=3,$ then $(x-3)(x-4)=0.$
If $x=2,$ then $(-1)(-2)=2.$
Either way, it is even, so both solutions are valid.
All solutions are $ 1,2,3,4$.
Hint:
$(-1)^k=1$ (if $k$ is an even integer).
In other words, you forgot the case that $x^2-5x+5=-1.$
Solving this, we get $x^2-5x+6=0rightarrow x=2,3.$
We check that the exponent is even.
If $x=3,$ then $(x-3)(x-4)=0.$
If $x=2,$ then $(-1)(-2)=2.$
Either way, it is even, so both solutions are valid.
All solutions are $ 1,2,3,4$.
edited Aug 31 at 8:43
tarit goswami
1,158219
1,158219
answered Aug 31 at 1:31
Jason Kim
53016
53016
So how would you use this to get the other possibilities? Sorry, I'm still quite confused.
â Julie
Aug 31 at 1:33
Solve for $x^2-5x+5=-1$ and check the exponent is even.
â Jason Kim
Aug 31 at 1:34
In particular ~ there are three possibilities: The exponent is $0,$ the base is $1/-1.$ $-1$ is the case that was not included in the solution.
â Jason Kim
Aug 31 at 1:37
1
@Julie You should actually answer this question yourself. Do you just have this equation or any more information? For example, is the question to find all integer values of $x$ solving the equation?
â amsmath
Aug 31 at 1:59
1
@JasonKim Recently in an entrance exam it was asked to explain why these are all solutions and there is no other(in first step asked to find the solutions). I have found all, but was not able to explain properly(I have explained by taking $log$ both side) why these are all solutions and there are no other. Can you explain :How we are sure there are no other solutions
?
â tarit goswami
Aug 31 at 4:57
 |Â
show 3 more comments
So how would you use this to get the other possibilities? Sorry, I'm still quite confused.
â Julie
Aug 31 at 1:33
Solve for $x^2-5x+5=-1$ and check the exponent is even.
â Jason Kim
Aug 31 at 1:34
In particular ~ there are three possibilities: The exponent is $0,$ the base is $1/-1.$ $-1$ is the case that was not included in the solution.
â Jason Kim
Aug 31 at 1:37
1
@Julie You should actually answer this question yourself. Do you just have this equation or any more information? For example, is the question to find all integer values of $x$ solving the equation?
â amsmath
Aug 31 at 1:59
1
@JasonKim Recently in an entrance exam it was asked to explain why these are all solutions and there is no other(in first step asked to find the solutions). I have found all, but was not able to explain properly(I have explained by taking $log$ both side) why these are all solutions and there are no other. Can you explain :How we are sure there are no other solutions
?
â tarit goswami
Aug 31 at 4:57
So how would you use this to get the other possibilities? Sorry, I'm still quite confused.
â Julie
Aug 31 at 1:33
So how would you use this to get the other possibilities? Sorry, I'm still quite confused.
â Julie
Aug 31 at 1:33
Solve for $x^2-5x+5=-1$ and check the exponent is even.
â Jason Kim
Aug 31 at 1:34
Solve for $x^2-5x+5=-1$ and check the exponent is even.
â Jason Kim
Aug 31 at 1:34
In particular ~ there are three possibilities: The exponent is $0,$ the base is $1/-1.$ $-1$ is the case that was not included in the solution.
â Jason Kim
Aug 31 at 1:37
In particular ~ there are three possibilities: The exponent is $0,$ the base is $1/-1.$ $-1$ is the case that was not included in the solution.
â Jason Kim
Aug 31 at 1:37
1
1
@Julie You should actually answer this question yourself. Do you just have this equation or any more information? For example, is the question to find all integer values of $x$ solving the equation?
â amsmath
Aug 31 at 1:59
@Julie You should actually answer this question yourself. Do you just have this equation or any more information? For example, is the question to find all integer values of $x$ solving the equation?
â amsmath
Aug 31 at 1:59
1
1
@JasonKim Recently in an entrance exam it was asked to explain why these are all solutions and there is no other(in first step asked to find the solutions). I have found all, but was not able to explain properly(I have explained by taking $log$ both side) why these are all solutions and there are no other. Can you explain :
How we are sure there are no other solutions
?â tarit goswami
Aug 31 at 4:57
@JasonKim Recently in an entrance exam it was asked to explain why these are all solutions and there is no other(in first step asked to find the solutions). I have found all, but was not able to explain properly(I have explained by taking $log$ both side) why these are all solutions and there are no other. Can you explain :
How we are sure there are no other solutions
?â tarit goswami
Aug 31 at 4:57
 |Â
show 3 more comments
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1
Please check the equation. The parentheses don't balance. (While you're editing, put the equation into the title of your question.)
â Blue
Aug 31 at 1:18
1
I have adjusted the problem's typesetting to be consistent with the $4,3,1$ solutions given.
â vadim123
Aug 31 at 1:29
$x^2-7x+12 = (x-3)(x-4)$
â amsmath
Aug 31 at 1:31
For positive $x^2-5x+5$ your solutions are the only ones. The question is whether negative $x^2-5x+5$ is allowed.
â amsmath
Aug 31 at 1:38
No, wait. For $x^2-5x+5 < 0$ for $x=3$.
â amsmath
Aug 31 at 1:52