Algebra polynomial
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if $P(x) = (2x-5)^2017+(2x-5)^2015+(x-4)^2017+(x-4)^2015+3x-9 = 0$ find the real roots. I found one real root $3$ by making sum of polynomial equating to o and finding the value of $x$ but not able to proceed further if any other root exit in real.
polynomials roots factoring
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up vote
1
down vote
favorite
if $P(x) = (2x-5)^2017+(2x-5)^2015+(x-4)^2017+(x-4)^2015+3x-9 = 0$ find the real roots. I found one real root $3$ by making sum of polynomial equating to o and finding the value of $x$ but not able to proceed further if any other root exit in real.
polynomials roots factoring
making sum of polynomial
â shirish
Aug 19 at 7:06
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
if $P(x) = (2x-5)^2017+(2x-5)^2015+(x-4)^2017+(x-4)^2015+3x-9 = 0$ find the real roots. I found one real root $3$ by making sum of polynomial equating to o and finding the value of $x$ but not able to proceed further if any other root exit in real.
polynomials roots factoring
if $P(x) = (2x-5)^2017+(2x-5)^2015+(x-4)^2017+(x-4)^2015+3x-9 = 0$ find the real roots. I found one real root $3$ by making sum of polynomial equating to o and finding the value of $x$ but not able to proceed further if any other root exit in real.
polynomials roots factoring
edited Aug 21 at 4:48
Michael Rozenberg
88.5k1579179
88.5k1579179
asked Aug 19 at 7:05
shirish
123
123
making sum of polynomial
â shirish
Aug 19 at 7:06
add a comment |Â
making sum of polynomial
â shirish
Aug 19 at 7:06
making sum of polynomial
â shirish
Aug 19 at 7:06
making sum of polynomial
â shirish
Aug 19 at 7:06
add a comment |Â
2 Answers
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It's $$3(x-3)left((2x-5)^2016-(2x-5)^2015(x-4)+...+(x-4)^2016+(2x-5)^2014-(2x-5)^2013(x-4)+...+(x-4)^2014+1right)=0.$$
Now, since the equation $y^2017=1$ has unique real root and the equation $y^2015=1$ has unique real root, we see that
$$y^2016-y^2015+...+1>0$$ and
$$y^2014-y^2013+...+1>0$$
and we see that can be $x-3=0$, which says that $3$ is an unique root of our equation.
add a comment |Â
up vote
2
down vote
Hint: calculate the derivative of $P$.
Hint2: $P'$ is positive in $mathbbR$.
can you Please explain? Do we need to use sturm's theorem
â shirish
Aug 19 at 7:55
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
It's $$3(x-3)left((2x-5)^2016-(2x-5)^2015(x-4)+...+(x-4)^2016+(2x-5)^2014-(2x-5)^2013(x-4)+...+(x-4)^2014+1right)=0.$$
Now, since the equation $y^2017=1$ has unique real root and the equation $y^2015=1$ has unique real root, we see that
$$y^2016-y^2015+...+1>0$$ and
$$y^2014-y^2013+...+1>0$$
and we see that can be $x-3=0$, which says that $3$ is an unique root of our equation.
add a comment |Â
up vote
2
down vote
It's $$3(x-3)left((2x-5)^2016-(2x-5)^2015(x-4)+...+(x-4)^2016+(2x-5)^2014-(2x-5)^2013(x-4)+...+(x-4)^2014+1right)=0.$$
Now, since the equation $y^2017=1$ has unique real root and the equation $y^2015=1$ has unique real root, we see that
$$y^2016-y^2015+...+1>0$$ and
$$y^2014-y^2013+...+1>0$$
and we see that can be $x-3=0$, which says that $3$ is an unique root of our equation.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It's $$3(x-3)left((2x-5)^2016-(2x-5)^2015(x-4)+...+(x-4)^2016+(2x-5)^2014-(2x-5)^2013(x-4)+...+(x-4)^2014+1right)=0.$$
Now, since the equation $y^2017=1$ has unique real root and the equation $y^2015=1$ has unique real root, we see that
$$y^2016-y^2015+...+1>0$$ and
$$y^2014-y^2013+...+1>0$$
and we see that can be $x-3=0$, which says that $3$ is an unique root of our equation.
It's $$3(x-3)left((2x-5)^2016-(2x-5)^2015(x-4)+...+(x-4)^2016+(2x-5)^2014-(2x-5)^2013(x-4)+...+(x-4)^2014+1right)=0.$$
Now, since the equation $y^2017=1$ has unique real root and the equation $y^2015=1$ has unique real root, we see that
$$y^2016-y^2015+...+1>0$$ and
$$y^2014-y^2013+...+1>0$$
and we see that can be $x-3=0$, which says that $3$ is an unique root of our equation.
edited Aug 19 at 7:31
answered Aug 19 at 7:25
Michael Rozenberg
88.5k1579179
88.5k1579179
add a comment |Â
add a comment |Â
up vote
2
down vote
Hint: calculate the derivative of $P$.
Hint2: $P'$ is positive in $mathbbR$.
can you Please explain? Do we need to use sturm's theorem
â shirish
Aug 19 at 7:55
add a comment |Â
up vote
2
down vote
Hint: calculate the derivative of $P$.
Hint2: $P'$ is positive in $mathbbR$.
can you Please explain? Do we need to use sturm's theorem
â shirish
Aug 19 at 7:55
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: calculate the derivative of $P$.
Hint2: $P'$ is positive in $mathbbR$.
Hint: calculate the derivative of $P$.
Hint2: $P'$ is positive in $mathbbR$.
edited Aug 19 at 8:15
answered Aug 19 at 7:30
ploosu2
4,065923
4,065923
can you Please explain? Do we need to use sturm's theorem
â shirish
Aug 19 at 7:55
add a comment |Â
can you Please explain? Do we need to use sturm's theorem
â shirish
Aug 19 at 7:55
can you Please explain? Do we need to use sturm's theorem
â shirish
Aug 19 at 7:55
can you Please explain? Do we need to use sturm's theorem
â shirish
Aug 19 at 7:55
add a comment |Â
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making sum of polynomial
â shirish
Aug 19 at 7:06