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Finding the sum of squares of the real roots
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Let $r_1,r_2,r_3,cdots,r_n$ be the distinct real zeroes of the equation $$x^8-14x^4-8x^3-x^2+1=0.$$ Then $r_1^2+r_2^2+r_3^2+cdots+r_n^2$ is $$(A),3quad(B),14quad(C),8quad(D),16$$
I can get the sum of the squares of all roots using Vieta’s formulae, but I don't know actually how to proceed in this question.
Do I need to draw a graph and then find the answers or is there a sum trick here which I am not able to see through?
polynomials roots
edited 18 mins ago
egreg
167k1180189
asked 1 hour ago
ashish deo singh
1485
add a comment |Â
up vote
6
down vote
favorite
Let $r_1,r_2,r_3,cdots,r_n$ be the distinct real zeroes of the equation $$x^8-14x^4-8x^3-x^2+1=0.$$ Then $r_1^2+r_2^2+r_3^2+cdots+r_n^2$ is $$(A),3quad(B),14quad(C),8quad(D),16$$
I can get the sum of the squares of all roots using Vieta’s formulae, but I don't know actually how to proceed in this question.
Do I need to draw a graph and then find the answers or is there a sum trick here which I am not able to see through?
polynomials roots
edited 18 mins ago
egreg
167k1180189
asked 1 hour ago
ashish deo singh
1485
Once you get Vieta's formulae you can continue with Newton's identity, basically $a_ns_2+a_n-1s_1+a_n-2=0$. But as dxiv said, you have first to isolate the polynomial with real roots.
– zwim
34 mins ago
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Let $r_1,r_2,r_3,cdots,r_n$ be the distinct real zeroes of the equation $$x^8-14x^4-8x^3-x^2+1=0.$$ Then $r_1^2+r_2^2+r_3^2+cdots+r_n^2$ is $$(A),3quad(B),14quad(C),8quad(D),16$$
I can get the sum of the squares of all roots using Vieta’s formulae, but I don't know actually how to proceed in this question.
Do I need to draw a graph and then find the answers or is there a sum trick here which I am not able to see through?
polynomials roots
edited 18 mins ago
egreg
167k1180189
asked 1 hour ago
ashish deo singh
1485
Let $r_1,r_2,r_3,cdots,r_n$ be the distinct real zeroes of the equation $$x^8-14x^4-8x^3-x^2+1=0.$$ Then $r_1^2+r_2^2+r_3^2+cdots+r_n^2$ is $$(A),3quad(B),14quad(C),8quad(D),16$$
I can get the sum of the squares of all roots using Vieta’s formulae, but I don't know actually how to proceed in this question.
Do I need to draw a graph and then find the answers or is there a sum trick here which I am not able to see through?
polynomials roots
polynomials roots
edited 18 mins ago
egreg
167k1180189
asked 1 hour ago
ashish deo singh
1485
edited 18 mins ago
egreg
167k1180189
asked 1 hour ago
ashish deo singh
1485
edited 18 mins ago
egreg
167k1180189
edited 18 mins ago
egreg
167k1180189
edited 18 mins ago
egreg
167k1180189
167k1180189
asked 1 hour ago
ashish deo singh
1485
asked 1 hour ago
ashish deo singh
1485
asked 1 hour ago
ashish deo singh
1485
1485
Once you get Vieta's formulae you can continue with Newton's identity, basically $a_ns_2+a_n-1s_1+a_n-2=0$. But as dxiv said, you have first to isolate the polynomial with real roots.
– zwim
34 mins ago
add a comment |Â
Once you get Vieta's formulae you can continue with Newton's identity, basically $a_ns_2+a_n-1s_1+a_n-2=0$. But as dxiv said, you have first to isolate the polynomial with real roots.
– zwim
34 mins ago
Once you get Vieta's formulae you can continue with Newton's identity, basically $a_ns_2+a_n-1s_1+a_n-2=0$. But as dxiv said, you have first to isolate the polynomial with real roots.
– zwim
34 mins ago
Once you get Vieta's formulae you can continue with Newton's identity, basically $a_ns_2+a_n-1s_1+a_n-2=0$. But as dxiv said, you have first to isolate the polynomial with real roots.
– zwim
34 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
Hint: Â the real roots are the roots of the first factor:
$$
beginalign
x^8colorred+2x^4-2x^4-14x^4-8x^3-x^2+1 &= (x^4+1)^2 - x^2(4x+1)^2 \
&= (x^4 - 4 x^2 - x+1)(x^4+4x^2+x+1)
endalign
$$
answered 1 hour ago
dxiv
56.1k64798
add a comment |Â
up vote
3
down vote
Let $$f(x)=x^8-14x^4-8x^3-x^2+1implies f'(x)=8x^7-56x^3-24x^2-2x=0$$ for critical points. Then clearly $x=0$ is one, and is a maximum since $f''(0)<0$; $f(0)=1$.
We get $$g(x)=4x^6-28x^3-12x-1=0$$ and since $g$ is continuous, by the Mean Value Theorem, there is a minimum between $(-0.2,0)$ and a maximum between $(-0.4,-0.2)$. At these $x$, $f(x)$ is above the $x$-axis.
Hence there are at most four roots left. In a similar way, we test at $0.2$ intervals, and we find that $$f(-1.8)>0,quad f(-1.6)<0\f(-0.8)<0,quad f(-0.6)>0\f(0.2)>0,quad f(0.4)<0\f(2)<0,quad f(2.2)>0.$$
Thus we have that $$(-1.6)^2+(-0.2)^2+0.2^2+2^2<r_1^2+r_2^2+r_3^2+r_4^2<(-1.8)^2+(-0.8)^2+0.4^2+2.2^2$$ or that $$6.64<r_1^2+r_2^2+r_3^2+r_4^2<8.88$$ so the only option must be $(C)$.
answered 1 hour ago
TheSimpliFire
10.9k62154
but this is too much of approximation and working through options. wont there be any other way to find exact ans
– ashish deo singh
1 hour ago
Yes, @dxiv has given a much neater one.
– TheSimpliFire
1 hour ago
add a comment |Â
up vote
0
down vote
Following the foundamental hint by dxiv, for $x^4-4x^2-x+1$ by Vieta's formula we have
- $S_1=sum r_i=-a_3=0$
- $S_2=sum r_ir_j=a_2=-4$
- $S_3=sum r_ir_jr_k=-a_1=1$
- $S_4=r_1r_2r_3r_4=a_0=1$
and by Newton's sums we have that
- $P_1=sum r_i=S_1=0$
- $P_2=sum r_i^2=S_1P_1-2S_2=8$
answered 23 mins ago
gimusi
72.7k73888
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Hint: Â the real roots are the roots of the first factor:
$$
beginalign
x^8colorred+2x^4-2x^4-14x^4-8x^3-x^2+1 &= (x^4+1)^2 - x^2(4x+1)^2 \
&= (x^4 - 4 x^2 - x+1)(x^4+4x^2+x+1)
endalign
$$
answered 1 hour ago
dxiv
56.1k64798
add a comment |Â
up vote
4
down vote
Hint: Â the real roots are the roots of the first factor:
$$
beginalign
x^8colorred+2x^4-2x^4-14x^4-8x^3-x^2+1 &= (x^4+1)^2 - x^2(4x+1)^2 \
&= (x^4 - 4 x^2 - x+1)(x^4+4x^2+x+1)
endalign
$$
answered 1 hour ago
dxiv
56.1k64798
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Hint: Â the real roots are the roots of the first factor:
$$
beginalign
x^8colorred+2x^4-2x^4-14x^4-8x^3-x^2+1 &= (x^4+1)^2 - x^2(4x+1)^2 \
&= (x^4 - 4 x^2 - x+1)(x^4+4x^2+x+1)
endalign
$$
answered 1 hour ago
dxiv
56.1k64798
Hint: Â the real roots are the roots of the first factor:
$$
beginalign
x^8colorred+2x^4-2x^4-14x^4-8x^3-x^2+1 &= (x^4+1)^2 - x^2(4x+1)^2 \
&= (x^4 - 4 x^2 - x+1)(x^4+4x^2+x+1)
endalign
$$
answered 1 hour ago
dxiv
56.1k64798
answered 1 hour ago
dxiv
56.1k64798
answered 1 hour ago
dxiv
56.1k64798
answered 1 hour ago
dxiv
56.1k64798
56.1k64798
add a comment |Â
add a comment |Â
up vote
3
down vote
Let $$f(x)=x^8-14x^4-8x^3-x^2+1implies f'(x)=8x^7-56x^3-24x^2-2x=0$$ for critical points. Then clearly $x=0$ is one, and is a maximum since $f''(0)<0$; $f(0)=1$.
We get $$g(x)=4x^6-28x^3-12x-1=0$$ and since $g$ is continuous, by the Mean Value Theorem, there is a minimum between $(-0.2,0)$ and a maximum between $(-0.4,-0.2)$. At these $x$, $f(x)$ is above the $x$-axis.
Hence there are at most four roots left. In a similar way, we test at $0.2$ intervals, and we find that $$f(-1.8)>0,quad f(-1.6)<0\f(-0.8)<0,quad f(-0.6)>0\f(0.2)>0,quad f(0.4)<0\f(2)<0,quad f(2.2)>0.$$
Thus we have that $$(-1.6)^2+(-0.2)^2+0.2^2+2^2<r_1^2+r_2^2+r_3^2+r_4^2<(-1.8)^2+(-0.8)^2+0.4^2+2.2^2$$ or that $$6.64<r_1^2+r_2^2+r_3^2+r_4^2<8.88$$ so the only option must be $(C)$.
answered 1 hour ago
TheSimpliFire
10.9k62154
but this is too much of approximation and working through options. wont there be any other way to find exact ans
– ashish deo singh
1 hour ago
Yes, @dxiv has given a much neater one.
– TheSimpliFire
1 hour ago
add a comment |Â
up vote
3
down vote
Let $$f(x)=x^8-14x^4-8x^3-x^2+1implies f'(x)=8x^7-56x^3-24x^2-2x=0$$ for critical points. Then clearly $x=0$ is one, and is a maximum since $f''(0)<0$; $f(0)=1$.
We get $$g(x)=4x^6-28x^3-12x-1=0$$ and since $g$ is continuous, by the Mean Value Theorem, there is a minimum between $(-0.2,0)$ and a maximum between $(-0.4,-0.2)$. At these $x$, $f(x)$ is above the $x$-axis.
Hence there are at most four roots left. In a similar way, we test at $0.2$ intervals, and we find that $$f(-1.8)>0,quad f(-1.6)<0\f(-0.8)<0,quad f(-0.6)>0\f(0.2)>0,quad f(0.4)<0\f(2)<0,quad f(2.2)>0.$$
Thus we have that $$(-1.6)^2+(-0.2)^2+0.2^2+2^2<r_1^2+r_2^2+r_3^2+r_4^2<(-1.8)^2+(-0.8)^2+0.4^2+2.2^2$$ or that $$6.64<r_1^2+r_2^2+r_3^2+r_4^2<8.88$$ so the only option must be $(C)$.
answered 1 hour ago
TheSimpliFire
10.9k62154
but this is too much of approximation and working through options. wont there be any other way to find exact ans
– ashish deo singh
1 hour ago
Yes, @dxiv has given a much neater one.
– TheSimpliFire
1 hour ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $$f(x)=x^8-14x^4-8x^3-x^2+1implies f'(x)=8x^7-56x^3-24x^2-2x=0$$ for critical points. Then clearly $x=0$ is one, and is a maximum since $f''(0)<0$; $f(0)=1$.
We get $$g(x)=4x^6-28x^3-12x-1=0$$ and since $g$ is continuous, by the Mean Value Theorem, there is a minimum between $(-0.2,0)$ and a maximum between $(-0.4,-0.2)$. At these $x$, $f(x)$ is above the $x$-axis.
Hence there are at most four roots left. In a similar way, we test at $0.2$ intervals, and we find that $$f(-1.8)>0,quad f(-1.6)<0\f(-0.8)<0,quad f(-0.6)>0\f(0.2)>0,quad f(0.4)<0\f(2)<0,quad f(2.2)>0.$$
Thus we have that $$(-1.6)^2+(-0.2)^2+0.2^2+2^2<r_1^2+r_2^2+r_3^2+r_4^2<(-1.8)^2+(-0.8)^2+0.4^2+2.2^2$$ or that $$6.64<r_1^2+r_2^2+r_3^2+r_4^2<8.88$$ so the only option must be $(C)$.
answered 1 hour ago
TheSimpliFire
10.9k62154
Let $$f(x)=x^8-14x^4-8x^3-x^2+1implies f'(x)=8x^7-56x^3-24x^2-2x=0$$ for critical points. Then clearly $x=0$ is one, and is a maximum since $f''(0)<0$; $f(0)=1$.
We get $$g(x)=4x^6-28x^3-12x-1=0$$ and since $g$ is continuous, by the Mean Value Theorem, there is a minimum between $(-0.2,0)$ and a maximum between $(-0.4,-0.2)$. At these $x$, $f(x)$ is above the $x$-axis.
Hence there are at most four roots left. In a similar way, we test at $0.2$ intervals, and we find that $$f(-1.8)>0,quad f(-1.6)<0\f(-0.8)<0,quad f(-0.6)>0\f(0.2)>0,quad f(0.4)<0\f(2)<0,quad f(2.2)>0.$$
Thus we have that $$(-1.6)^2+(-0.2)^2+0.2^2+2^2<r_1^2+r_2^2+r_3^2+r_4^2<(-1.8)^2+(-0.8)^2+0.4^2+2.2^2$$ or that $$6.64<r_1^2+r_2^2+r_3^2+r_4^2<8.88$$ so the only option must be $(C)$.
answered 1 hour ago
TheSimpliFire
10.9k62154
answered 1 hour ago
TheSimpliFire
10.9k62154
answered 1 hour ago
TheSimpliFire
10.9k62154
answered 1 hour ago
TheSimpliFire
10.9k62154
10.9k62154
but this is too much of approximation and working through options. wont there be any other way to find exact ans
– ashish deo singh
1 hour ago
Yes, @dxiv has given a much neater one.
– TheSimpliFire
1 hour ago
add a comment |Â
but this is too much of approximation and working through options. wont there be any other way to find exact ans
– ashish deo singh
1 hour ago
Yes, @dxiv has given a much neater one.
– TheSimpliFire
1 hour ago
but this is too much of approximation and working through options. wont there be any other way to find exact ans
– ashish deo singh
1 hour ago
but this is too much of approximation and working through options. wont there be any other way to find exact ans
– ashish deo singh
1 hour ago
Yes, @dxiv has given a much neater one.
– TheSimpliFire
1 hour ago
Yes, @dxiv has given a much neater one.
– TheSimpliFire
1 hour ago
add a comment |Â
up vote
0
down vote
Following the foundamental hint by dxiv, for $x^4-4x^2-x+1$ by Vieta's formula we have
- $S_1=sum r_i=-a_3=0$
- $S_2=sum r_ir_j=a_2=-4$
- $S_3=sum r_ir_jr_k=-a_1=1$
- $S_4=r_1r_2r_3r_4=a_0=1$
and by Newton's sums we have that
- $P_1=sum r_i=S_1=0$
- $P_2=sum r_i^2=S_1P_1-2S_2=8$
answered 23 mins ago
gimusi
72.7k73888
add a comment |Â
up vote
0
down vote
Following the foundamental hint by dxiv, for $x^4-4x^2-x+1$ by Vieta's formula we have
- $S_1=sum r_i=-a_3=0$
- $S_2=sum r_ir_j=a_2=-4$
- $S_3=sum r_ir_jr_k=-a_1=1$
- $S_4=r_1r_2r_3r_4=a_0=1$
and by Newton's sums we have that
- $P_1=sum r_i=S_1=0$
- $P_2=sum r_i^2=S_1P_1-2S_2=8$
answered 23 mins ago
gimusi
72.7k73888
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Following the foundamental hint by dxiv, for $x^4-4x^2-x+1$ by Vieta's formula we have
- $S_1=sum r_i=-a_3=0$
- $S_2=sum r_ir_j=a_2=-4$
- $S_3=sum r_ir_jr_k=-a_1=1$
- $S_4=r_1r_2r_3r_4=a_0=1$
and by Newton's sums we have that
- $P_1=sum r_i=S_1=0$
- $P_2=sum r_i^2=S_1P_1-2S_2=8$
answered 23 mins ago
gimusi
72.7k73888
Following the foundamental hint by dxiv, for $x^4-4x^2-x+1$ by Vieta's formula we have
- $S_1=sum r_i=-a_3=0$
- $S_2=sum r_ir_j=a_2=-4$
- $S_3=sum r_ir_jr_k=-a_1=1$
- $S_4=r_1r_2r_3r_4=a_0=1$
and by Newton's sums we have that
- $P_1=sum r_i=S_1=0$
- $P_2=sum r_i^2=S_1P_1-2S_2=8$
answered 23 mins ago
gimusi
72.7k73888
answered 23 mins ago
gimusi
72.7k73888
answered 23 mins ago
gimusi
72.7k73888
answered 23 mins ago
gimusi
72.7k73888
72.7k73888
add a comment |Â
add a comment |Â
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Once you get Vieta's formulae you can continue with Newton's identity, basically $a_ns_2+a_n-1s_1+a_n-2=0$. But as dxiv said, you have first to isolate the polynomial with real roots.
– zwim
34 mins ago