Mixing
Solution to the equation of a polynomial raised to the power of a polynomial.
Clash Royale CLAN TAG#URR8PPP
up vote
7
down vote
favorite
The problem at hand is, find the solutions of $x$ in the following equation:
$$ (x^2−7x+11)^x^2−7x+6=1 $$
My friend who gave me this questions, told me that you can find $6$ solutions without needing to graph the equation.
My approach was this: Use factoring and the fact that $z^0=1$ for $z≠0$ and $1^z=1$ for any $z$.
Factorising the exponent, we have:
$$ x^2-7x+6 = (x-1)(x-6) $$
Therefore, by making the exponent = 0, we have possible solutions as $x in 1,6 $
Making the base of the exponent = $1$, we get $$ x^2-7x+10 = 0 $$
$$ (x-2)(x-5)$$
Hence we can say $x in 2, 5 $.
However, I am unable to compute the last two solutions. Could anyone shed some light on how to proceed?
algebra-precalculus polynomials systems-of-equations
asked 48 mins ago
Tusky
321112
add a comment |Â
up vote
7
down vote
favorite
The problem at hand is, find the solutions of $x$ in the following equation:
$$ (x^2−7x+11)^x^2−7x+6=1 $$
My friend who gave me this questions, told me that you can find $6$ solutions without needing to graph the equation.
My approach was this: Use factoring and the fact that $z^0=1$ for $z≠0$ and $1^z=1$ for any $z$.
Factorising the exponent, we have:
$$ x^2-7x+6 = (x-1)(x-6) $$
Therefore, by making the exponent = 0, we have possible solutions as $x in 1,6 $
Making the base of the exponent = $1$, we get $$ x^2-7x+10 = 0 $$
$$ (x-2)(x-5)$$
Hence we can say $x in 2, 5 $.
However, I am unable to compute the last two solutions. Could anyone shed some light on how to proceed?
algebra-precalculus polynomials systems-of-equations
asked 48 mins ago
Tusky
321112
There are no more solutions. Probably your friend thought about the number $6$ as one solution seen immediately.
– Berci
41 mins ago
1
Hint: If the base is $-1$, and the exponent is an even integer, then . . .
– quasi
36 mins ago
@Berci: There are more solutions; take a look at the answers.
– user21820
5 mins ago
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
The problem at hand is, find the solutions of $x$ in the following equation:
$$ (x^2−7x+11)^x^2−7x+6=1 $$
My friend who gave me this questions, told me that you can find $6$ solutions without needing to graph the equation.
My approach was this: Use factoring and the fact that $z^0=1$ for $z≠0$ and $1^z=1$ for any $z$.
Factorising the exponent, we have:
$$ x^2-7x+6 = (x-1)(x-6) $$
Therefore, by making the exponent = 0, we have possible solutions as $x in 1,6 $
Making the base of the exponent = $1$, we get $$ x^2-7x+10 = 0 $$
$$ (x-2)(x-5)$$
Hence we can say $x in 2, 5 $.
However, I am unable to compute the last two solutions. Could anyone shed some light on how to proceed?
algebra-precalculus polynomials systems-of-equations
asked 48 mins ago
Tusky
321112
The problem at hand is, find the solutions of $x$ in the following equation:
$$ (x^2−7x+11)^x^2−7x+6=1 $$
My friend who gave me this questions, told me that you can find $6$ solutions without needing to graph the equation.
My approach was this: Use factoring and the fact that $z^0=1$ for $z≠0$ and $1^z=1$ for any $z$.
Factorising the exponent, we have:
$$ x^2-7x+6 = (x-1)(x-6) $$
Therefore, by making the exponent = 0, we have possible solutions as $x in 1,6 $
Making the base of the exponent = $1$, we get $$ x^2-7x+10 = 0 $$
$$ (x-2)(x-5)$$
Hence we can say $x in 2, 5 $.
However, I am unable to compute the last two solutions. Could anyone shed some light on how to proceed?
algebra-precalculus polynomials systems-of-equations
algebra-precalculus polynomials systems-of-equations
asked 48 mins ago
Tusky
321112
asked 48 mins ago
Tusky
321112
asked 48 mins ago
Tusky
321112
asked 48 mins ago
Tusky
321112
asked 48 mins ago
Tusky
321112
321112
There are no more solutions. Probably your friend thought about the number $6$ as one solution seen immediately.
– Berci
41 mins ago
1
Hint: If the base is $-1$, and the exponent is an even integer, then . . .
– quasi
36 mins ago
@Berci: There are more solutions; take a look at the answers.
– user21820
5 mins ago
add a comment |Â
There are no more solutions. Probably your friend thought about the number $6$ as one solution seen immediately.
– Berci
41 mins ago
1
Hint: If the base is $-1$, and the exponent is an even integer, then . . .
– quasi
36 mins ago
@Berci: There are more solutions; take a look at the answers.
– user21820
5 mins ago
There are no more solutions. Probably your friend thought about the number $6$ as one solution seen immediately.
– Berci
41 mins ago
There are no more solutions. Probably your friend thought about the number $6$ as one solution seen immediately.
– Berci
41 mins ago
1
1
Hint: If the base is $-1$, and the exponent is an even integer, then . . .
– quasi
36 mins ago
Hint: If the base is $-1$, and the exponent is an even integer, then . . .
– quasi
36 mins ago
@Berci: There are more solutions; take a look at the answers.
– user21820
5 mins ago
@Berci: There are more solutions; take a look at the answers.
– user21820
5 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
The only venue you can explore then is to see that $1$ is the power of one other number, that is $-1$.
$(-1)^2k=1$, $forall kin mathbbZ$
By stating that, you can see that if you have both (you need to fumble around a bit and do a bit of trial and error):
$x^2-7x+11=-1$ AND $x^2-7x+6=-6$, you would have then $(-1)^-6=frac1(-1)^6=1$
And of course both equations are in fact the same (otherwise you would not be able to find solutions, that is:
$x^2-7x+12=(x-3)(x-4)=0$ with solutions $(3,4)$.
So $(1,2,3,4,5,6)$ are the six solutions.
answered 35 mins ago
Martigan
5,037916
add a comment |Â
up vote
5
down vote
Denote $a=x^2-7x+11.$ The equation becomes $a^a-5=1,$ or equivalently $$a^a=a^5,$$ which has in $mathbbR$ the solutions $ain 5,1,-1.$ Solving the corresponding quadratic equations we get the solutions $xin 1,6,2,5,3,4.$
answered 18 mins ago
user376343
635311
3
Very elegant way of presenting the problem. In my opinion better than my own answer. I upvote!
– Martigan
11 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
The only venue you can explore then is to see that $1$ is the power of one other number, that is $-1$.
$(-1)^2k=1$, $forall kin mathbbZ$
By stating that, you can see that if you have both (you need to fumble around a bit and do a bit of trial and error):
$x^2-7x+11=-1$ AND $x^2-7x+6=-6$, you would have then $(-1)^-6=frac1(-1)^6=1$
And of course both equations are in fact the same (otherwise you would not be able to find solutions, that is:
$x^2-7x+12=(x-3)(x-4)=0$ with solutions $(3,4)$.
So $(1,2,3,4,5,6)$ are the six solutions.
answered 35 mins ago
Martigan
5,037916
add a comment |Â
up vote
5
down vote
The only venue you can explore then is to see that $1$ is the power of one other number, that is $-1$.
$(-1)^2k=1$, $forall kin mathbbZ$
By stating that, you can see that if you have both (you need to fumble around a bit and do a bit of trial and error):
$x^2-7x+11=-1$ AND $x^2-7x+6=-6$, you would have then $(-1)^-6=frac1(-1)^6=1$
And of course both equations are in fact the same (otherwise you would not be able to find solutions, that is:
$x^2-7x+12=(x-3)(x-4)=0$ with solutions $(3,4)$.
So $(1,2,3,4,5,6)$ are the six solutions.
answered 35 mins ago
Martigan
5,037916
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The only venue you can explore then is to see that $1$ is the power of one other number, that is $-1$.
$(-1)^2k=1$, $forall kin mathbbZ$
By stating that, you can see that if you have both (you need to fumble around a bit and do a bit of trial and error):
$x^2-7x+11=-1$ AND $x^2-7x+6=-6$, you would have then $(-1)^-6=frac1(-1)^6=1$
And of course both equations are in fact the same (otherwise you would not be able to find solutions, that is:
$x^2-7x+12=(x-3)(x-4)=0$ with solutions $(3,4)$.
So $(1,2,3,4,5,6)$ are the six solutions.
answered 35 mins ago
Martigan
5,037916
The only venue you can explore then is to see that $1$ is the power of one other number, that is $-1$.
$(-1)^2k=1$, $forall kin mathbbZ$
By stating that, you can see that if you have both (you need to fumble around a bit and do a bit of trial and error):
$x^2-7x+11=-1$ AND $x^2-7x+6=-6$, you would have then $(-1)^-6=frac1(-1)^6=1$
And of course both equations are in fact the same (otherwise you would not be able to find solutions, that is:
$x^2-7x+12=(x-3)(x-4)=0$ with solutions $(3,4)$.
So $(1,2,3,4,5,6)$ are the six solutions.
answered 35 mins ago
Martigan
5,037916
answered 35 mins ago
Martigan
5,037916
answered 35 mins ago
Martigan
5,037916
answered 35 mins ago
Martigan
5,037916
5,037916
add a comment |Â
add a comment |Â
up vote
5
down vote
Denote $a=x^2-7x+11.$ The equation becomes $a^a-5=1,$ or equivalently $$a^a=a^5,$$ which has in $mathbbR$ the solutions $ain 5,1,-1.$ Solving the corresponding quadratic equations we get the solutions $xin 1,6,2,5,3,4.$
answered 18 mins ago
user376343
635311
3
Very elegant way of presenting the problem. In my opinion better than my own answer. I upvote!
– Martigan
11 mins ago
add a comment |Â
up vote
5
down vote
Denote $a=x^2-7x+11.$ The equation becomes $a^a-5=1,$ or equivalently $$a^a=a^5,$$ which has in $mathbbR$ the solutions $ain 5,1,-1.$ Solving the corresponding quadratic equations we get the solutions $xin 1,6,2,5,3,4.$
answered 18 mins ago
user376343
635311
3
Very elegant way of presenting the problem. In my opinion better than my own answer. I upvote!
– Martigan
11 mins ago
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Denote $a=x^2-7x+11.$ The equation becomes $a^a-5=1,$ or equivalently $$a^a=a^5,$$ which has in $mathbbR$ the solutions $ain 5,1,-1.$ Solving the corresponding quadratic equations we get the solutions $xin 1,6,2,5,3,4.$
answered 18 mins ago
user376343
635311
Denote $a=x^2-7x+11.$ The equation becomes $a^a-5=1,$ or equivalently $$a^a=a^5,$$ which has in $mathbbR$ the solutions $ain 5,1,-1.$ Solving the corresponding quadratic equations we get the solutions $xin 1,6,2,5,3,4.$
answered 18 mins ago
user376343
635311
answered 18 mins ago
user376343
635311
answered 18 mins ago
user376343
635311
answered 18 mins ago
user376343
635311
635311
3
Very elegant way of presenting the problem. In my opinion better than my own answer. I upvote!
– Martigan
11 mins ago
add a comment |Â
3
Very elegant way of presenting the problem. In my opinion better than my own answer. I upvote!
– Martigan
11 mins ago
3
3
Very elegant way of presenting the problem. In my opinion better than my own answer. I upvote!
– Martigan
11 mins ago
Very elegant way of presenting the problem. In my opinion better than my own answer. I upvote!
– Martigan
11 mins ago
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2943102%2fsolution-to-the-equation-of-a-polynomial-raised-to-the-power-of-a-polynomial%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
There are no more solutions. Probably your friend thought about the number $6$ as one solution seen immediately.
– Berci
41 mins ago
1
Hint: If the base is $-1$, and the exponent is an even integer, then . . .
– quasi
36 mins ago
@Berci: There are more solutions; take a look at the answers.
– user21820
5 mins ago