Evaluating the modulus of roots of a 6 degree polynomial
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Let $x_1,x_2,x_3,x_4,x_5,x_6$ be the roots of the polynomial equation
$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0.$$
Then,
(A)$|x_i|=2$ for exactly one value of $i$
(B)$|x_i|=2$ for exactly two values of $i$
(C)$|x_i|=2$ for all values of $i$.
(D)$|x_i|=2$ for no value of $i$.
My attempt: I noticed that the polynomial forms a geometric
progression with common ratio $2/x$.
Then I summed the terms and found this relation $x^7=128$ which gives us
$x=2$ but $x=2$ doesn't satisfy the original polynomial.
I looked at the graph too and found that the polynomial has no real roots .
So now , how to find the modulus of the roots ?
polynomials roots
edited Aug 21 at 12:18
Clayton
18.3k22883
asked Aug 21 at 12:15
Alphanerd
8719
add a comment |Â
up vote
5
down vote
favorite
Let $x_1,x_2,x_3,x_4,x_5,x_6$ be the roots of the polynomial equation
$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0.$$
Then,
(A)$|x_i|=2$ for exactly one value of $i$
(B)$|x_i|=2$ for exactly two values of $i$
(C)$|x_i|=2$ for all values of $i$.
(D)$|x_i|=2$ for no value of $i$.
My attempt: I noticed that the polynomial forms a geometric
progression with common ratio $2/x$.
Then I summed the terms and found this relation $x^7=128$ which gives us
$x=2$ but $x=2$ doesn't satisfy the original polynomial.
I looked at the graph too and found that the polynomial has no real roots .
So now , how to find the modulus of the roots ?
polynomials roots
edited Aug 21 at 12:18
Clayton
18.3k22883
asked Aug 21 at 12:15
Alphanerd
8719
1
You found $x^ 7 = 128$. Note that $x neq 2$, of course. It turns out that, in fact, the above polynomial is a factor of $x^7 - 128$. Consequently, every root must have modulus equal to $2$, and we can write down all the roots explicitly using the seventh roots of unity.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 21 at 12:21
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let $x_1,x_2,x_3,x_4,x_5,x_6$ be the roots of the polynomial equation
$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0.$$
Then,
(A)$|x_i|=2$ for exactly one value of $i$
(B)$|x_i|=2$ for exactly two values of $i$
(C)$|x_i|=2$ for all values of $i$.
(D)$|x_i|=2$ for no value of $i$.
My attempt: I noticed that the polynomial forms a geometric
progression with common ratio $2/x$.
Then I summed the terms and found this relation $x^7=128$ which gives us
$x=2$ but $x=2$ doesn't satisfy the original polynomial.
I looked at the graph too and found that the polynomial has no real roots .
So now , how to find the modulus of the roots ?
polynomials roots
edited Aug 21 at 12:18
Clayton
18.3k22883
asked Aug 21 at 12:15
Alphanerd
8719
Let $x_1,x_2,x_3,x_4,x_5,x_6$ be the roots of the polynomial equation
$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0.$$
Then,
(A)$|x_i|=2$ for exactly one value of $i$
(B)$|x_i|=2$ for exactly two values of $i$
(C)$|x_i|=2$ for all values of $i$.
(D)$|x_i|=2$ for no value of $i$.
My attempt: I noticed that the polynomial forms a geometric
progression with common ratio $2/x$.
Then I summed the terms and found this relation $x^7=128$ which gives us
$x=2$ but $x=2$ doesn't satisfy the original polynomial.
I looked at the graph too and found that the polynomial has no real roots .
So now , how to find the modulus of the roots ?
polynomials roots
polynomials roots
edited Aug 21 at 12:18
Clayton
18.3k22883
asked Aug 21 at 12:15
Alphanerd
8719
edited Aug 21 at 12:18
Clayton
18.3k22883
asked Aug 21 at 12:15
Alphanerd
8719
edited Aug 21 at 12:18
Clayton
18.3k22883
edited Aug 21 at 12:18
Clayton
18.3k22883
edited Aug 21 at 12:18
Clayton
18.3k22883
18.3k22883
asked Aug 21 at 12:15
Alphanerd
8719
asked Aug 21 at 12:15
Alphanerd
8719
asked Aug 21 at 12:15
Alphanerd
8719
8719
1
You found $x^ 7 = 128$. Note that $x neq 2$, of course. It turns out that, in fact, the above polynomial is a factor of $x^7 - 128$. Consequently, every root must have modulus equal to $2$, and we can write down all the roots explicitly using the seventh roots of unity.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 21 at 12:21
add a comment |Â
1
You found $x^ 7 = 128$. Note that $x neq 2$, of course. It turns out that, in fact, the above polynomial is a factor of $x^7 - 128$. Consequently, every root must have modulus equal to $2$, and we can write down all the roots explicitly using the seventh roots of unity.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 21 at 12:21
1
1
You found $x^ 7 = 128$. Note that $x neq 2$, of course. It turns out that, in fact, the above polynomial is a factor of $x^7 - 128$. Consequently, every root must have modulus equal to $2$, and we can write down all the roots explicitly using the seventh roots of unity.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 21 at 12:21
You found $x^ 7 = 128$. Note that $x neq 2$, of course. It turns out that, in fact, the above polynomial is a factor of $x^7 - 128$. Consequently, every root must have modulus equal to $2$, and we can write down all the roots explicitly using the seventh roots of unity.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 21 at 12:21
add a comment |Â
2 Answers
2
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oldest
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up vote
4
down vote
accepted
$$f(x)=sum_r=0^6(x^r2^6-r)=2^6dfracleft(dfrac x2right)^7-1dfrac x2-1=dfracx^7-128x-2$$
Clearly, $f(2)ne0$
$implies x^7=128$ with $xne2$
$iffleft(dfrac x2right)^7=1$
Now take modulus in both sides using $|z^m|=|z|^m$
answered Aug 21 at 12:19
lab bhattacharjee
216k14153266
add a comment |Â
up vote
3
down vote
As $x=2$ is not a root, the equation is
$$fracx^7-128x-2=0.$$
From this we draw
$$|x^7|=|x|^7=128$$ and the only real solution $$|x|=2.$$
answered Aug 21 at 12:20
Yves Daoust
115k666209
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
$$f(x)=sum_r=0^6(x^r2^6-r)=2^6dfracleft(dfrac x2right)^7-1dfrac x2-1=dfracx^7-128x-2$$
Clearly, $f(2)ne0$
$implies x^7=128$ with $xne2$
$iffleft(dfrac x2right)^7=1$
Now take modulus in both sides using $|z^m|=|z|^m$
answered Aug 21 at 12:19
lab bhattacharjee
216k14153266
add a comment |Â
up vote
4
down vote
accepted
$$f(x)=sum_r=0^6(x^r2^6-r)=2^6dfracleft(dfrac x2right)^7-1dfrac x2-1=dfracx^7-128x-2$$
Clearly, $f(2)ne0$
$implies x^7=128$ with $xne2$
$iffleft(dfrac x2right)^7=1$
Now take modulus in both sides using $|z^m|=|z|^m$
answered Aug 21 at 12:19
lab bhattacharjee
216k14153266
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
$$f(x)=sum_r=0^6(x^r2^6-r)=2^6dfracleft(dfrac x2right)^7-1dfrac x2-1=dfracx^7-128x-2$$
Clearly, $f(2)ne0$
$implies x^7=128$ with $xne2$
$iffleft(dfrac x2right)^7=1$
Now take modulus in both sides using $|z^m|=|z|^m$
answered Aug 21 at 12:19
lab bhattacharjee
216k14153266
$$f(x)=sum_r=0^6(x^r2^6-r)=2^6dfracleft(dfrac x2right)^7-1dfrac x2-1=dfracx^7-128x-2$$
Clearly, $f(2)ne0$
$implies x^7=128$ with $xne2$
$iffleft(dfrac x2right)^7=1$
Now take modulus in both sides using $|z^m|=|z|^m$
answered Aug 21 at 12:19
lab bhattacharjee
216k14153266
answered Aug 21 at 12:19
lab bhattacharjee
216k14153266
answered Aug 21 at 12:19
lab bhattacharjee
216k14153266
answered Aug 21 at 12:19
lab bhattacharjee
216k14153266
216k14153266
add a comment |Â
add a comment |Â
up vote
3
down vote
As $x=2$ is not a root, the equation is
$$fracx^7-128x-2=0.$$
From this we draw
$$|x^7|=|x|^7=128$$ and the only real solution $$|x|=2.$$
answered Aug 21 at 12:20
Yves Daoust
115k666209
add a comment |Â
up vote
3
down vote
As $x=2$ is not a root, the equation is
$$fracx^7-128x-2=0.$$
From this we draw
$$|x^7|=|x|^7=128$$ and the only real solution $$|x|=2.$$
answered Aug 21 at 12:20
Yves Daoust
115k666209
add a comment |Â
up vote
3
down vote
up vote
3
down vote
As $x=2$ is not a root, the equation is
$$fracx^7-128x-2=0.$$
From this we draw
$$|x^7|=|x|^7=128$$ and the only real solution $$|x|=2.$$
answered Aug 21 at 12:20
Yves Daoust
115k666209
As $x=2$ is not a root, the equation is
$$fracx^7-128x-2=0.$$
From this we draw
$$|x^7|=|x|^7=128$$ and the only real solution $$|x|=2.$$
answered Aug 21 at 12:20
Yves Daoust
115k666209
answered Aug 21 at 12:20
Yves Daoust
115k666209
answered Aug 21 at 12:20
Yves Daoust
115k666209
answered Aug 21 at 12:20
Yves Daoust
115k666209
115k666209
add a comment |Â
add a comment |Â
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1
You found $x^ 7 = 128$. Note that $x neq 2$, of course. It turns out that, in fact, the above polynomial is a factor of $x^7 - 128$. Consequently, every root must have modulus equal to $2$, and we can write down all the roots explicitly using the seventh roots of unity.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 21 at 12:21