Mixing
Relation between Roots and Coefficients Question [closed]
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If the roots of the equation $x^3+px^2+qx+r=0$ are consecutive terms of a geometric series, prove that $q^3 = p^3r$.
Show that this condition is satisfied for the equation $8x^3-100x^2+250x-125=0$ and solve this equation.
polynomials
asked Jan 22 at 23:17
nicknick
334
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closed as off-topic by amd, Eric Wofsey, Alexander Gruber♦ Jan 23 at 1:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amd, Eric Wofsey, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
If the roots of the equation $x^3+px^2+qx+r=0$ are consecutive terms of a geometric series, prove that $q^3 = p^3r$.
Show that this condition is satisfied for the equation $8x^3-100x^2+250x-125=0$ and solve this equation.
polynomials
asked Jan 22 at 23:17
nicknick
334
$endgroup$
closed as off-topic by amd, Eric Wofsey, Alexander Gruber♦ Jan 23 at 1:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amd, Eric Wofsey, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
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Hint: either factor the equation or use Vieta’s formulas.
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– amd
Jan 23 at 0:03
add a comment |
$begingroup$
If the roots of the equation $x^3+px^2+qx+r=0$ are consecutive terms of a geometric series, prove that $q^3 = p^3r$.
Show that this condition is satisfied for the equation $8x^3-100x^2+250x-125=0$ and solve this equation.
polynomials
asked Jan 22 at 23:17
nicknick
334
$endgroup$
If the roots of the equation $x^3+px^2+qx+r=0$ are consecutive terms of a geometric series, prove that $q^3 = p^3r$.
Show that this condition is satisfied for the equation $8x^3-100x^2+250x-125=0$ and solve this equation.
polynomials
polynomials
asked Jan 22 at 23:17
nicknick
334
asked Jan 22 at 23:17
nicknick
334
asked Jan 22 at 23:17
nicknick
334
asked Jan 22 at 23:17
nicknick
334
asked Jan 22 at 23:17
nicknick
334
334
closed as off-topic by amd, Eric Wofsey, Alexander Gruber♦ Jan 23 at 1:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amd, Eric Wofsey, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amd, Eric Wofsey, Alexander Gruber♦ Jan 23 at 1:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amd, Eric Wofsey, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
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Hint: either factor the equation or use Vieta’s formulas.
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– amd
Jan 23 at 0:03
add a comment |
$begingroup$
Hint: either factor the equation or use Vieta’s formulas.
$endgroup$
– amd
Jan 23 at 0:03
$begingroup$
Hint: either factor the equation or use Vieta’s formulas.
$endgroup$
– amd
Jan 23 at 0:03
$begingroup$
Hint: either factor the equation or use Vieta’s formulas.
$endgroup$
– amd
Jan 23 at 0:03
add a comment |
1 Answer
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The roots can be written as $a, ab, ab^2$. Then $(x-a)(x-ab)(x-ab^2) = x^3 + px^2 + qx +r$. Multiplying the left hand side and comparing yields $p = -a(1+b+b^2), q = a^2b(1+b+b^2), r = -a^3b^3$. This shows that $q^3 = p^3r$.
Since $r = -a^3b^3$, we get $ab = -sqrt[3]{r}$ which is one of the three roots. Thus, if $q^3 = p^3r$, then it is worth to check whether $-sqrt[3]{r}$ is a root.
Note, however, that $q^3 = p^3r$ is not sufficient in order that the roots have the above form. Take for example $q = r = 0$ and $p ne 0$. Then the polynomial has a zero of order two at $0$ and a zero of order one at $-p$.
For the special equation we have $(100/8)^3 125/8 = (5^2 2^2/2^3)^3 5^3/2^3 = 5^9/2^6 =5^9 2^3/2^9 = (5^3 2/2^3)^3 = (250/8)^3$.
This means that we should consider $-sqrt[3]{r} = 5/2$. In fact, this is a root. Polynomial divison produces a quadratric equation with solutions $5(2 - sqrt{3})/2$ and $5(2 + sqrt{3})/2$. Hence we get $a = 5(2 - sqrt{3})/2$ and $b = 2 + sqrt{3}$.
answered Jan 23 at 1:16
Paul FrostPaul Frost
11.6k3934
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The roots can be written as $a, ab, ab^2$. Then $(x-a)(x-ab)(x-ab^2) = x^3 + px^2 + qx +r$. Multiplying the left hand side and comparing yields $p = -a(1+b+b^2), q = a^2b(1+b+b^2), r = -a^3b^3$. This shows that $q^3 = p^3r$.
Since $r = -a^3b^3$, we get $ab = -sqrt[3]{r}$ which is one of the three roots. Thus, if $q^3 = p^3r$, then it is worth to check whether $-sqrt[3]{r}$ is a root.
Note, however, that $q^3 = p^3r$ is not sufficient in order that the roots have the above form. Take for example $q = r = 0$ and $p ne 0$. Then the polynomial has a zero of order two at $0$ and a zero of order one at $-p$.
For the special equation we have $(100/8)^3 125/8 = (5^2 2^2/2^3)^3 5^3/2^3 = 5^9/2^6 =5^9 2^3/2^9 = (5^3 2/2^3)^3 = (250/8)^3$.
This means that we should consider $-sqrt[3]{r} = 5/2$. In fact, this is a root. Polynomial divison produces a quadratric equation with solutions $5(2 - sqrt{3})/2$ and $5(2 + sqrt{3})/2$. Hence we get $a = 5(2 - sqrt{3})/2$ and $b = 2 + sqrt{3}$.
answered Jan 23 at 1:16
Paul FrostPaul Frost
11.6k3934
$endgroup$
add a comment |
$begingroup$
The roots can be written as $a, ab, ab^2$. Then $(x-a)(x-ab)(x-ab^2) = x^3 + px^2 + qx +r$. Multiplying the left hand side and comparing yields $p = -a(1+b+b^2), q = a^2b(1+b+b^2), r = -a^3b^3$. This shows that $q^3 = p^3r$.
Since $r = -a^3b^3$, we get $ab = -sqrt[3]{r}$ which is one of the three roots. Thus, if $q^3 = p^3r$, then it is worth to check whether $-sqrt[3]{r}$ is a root.
Note, however, that $q^3 = p^3r$ is not sufficient in order that the roots have the above form. Take for example $q = r = 0$ and $p ne 0$. Then the polynomial has a zero of order two at $0$ and a zero of order one at $-p$.
For the special equation we have $(100/8)^3 125/8 = (5^2 2^2/2^3)^3 5^3/2^3 = 5^9/2^6 =5^9 2^3/2^9 = (5^3 2/2^3)^3 = (250/8)^3$.
This means that we should consider $-sqrt[3]{r} = 5/2$. In fact, this is a root. Polynomial divison produces a quadratric equation with solutions $5(2 - sqrt{3})/2$ and $5(2 + sqrt{3})/2$. Hence we get $a = 5(2 - sqrt{3})/2$ and $b = 2 + sqrt{3}$.
answered Jan 23 at 1:16
Paul FrostPaul Frost
11.6k3934
$endgroup$
add a comment |
$begingroup$
The roots can be written as $a, ab, ab^2$. Then $(x-a)(x-ab)(x-ab^2) = x^3 + px^2 + qx +r$. Multiplying the left hand side and comparing yields $p = -a(1+b+b^2), q = a^2b(1+b+b^2), r = -a^3b^3$. This shows that $q^3 = p^3r$.
Since $r = -a^3b^3$, we get $ab = -sqrt[3]{r}$ which is one of the three roots. Thus, if $q^3 = p^3r$, then it is worth to check whether $-sqrt[3]{r}$ is a root.
Note, however, that $q^3 = p^3r$ is not sufficient in order that the roots have the above form. Take for example $q = r = 0$ and $p ne 0$. Then the polynomial has a zero of order two at $0$ and a zero of order one at $-p$.
For the special equation we have $(100/8)^3 125/8 = (5^2 2^2/2^3)^3 5^3/2^3 = 5^9/2^6 =5^9 2^3/2^9 = (5^3 2/2^3)^3 = (250/8)^3$.
This means that we should consider $-sqrt[3]{r} = 5/2$. In fact, this is a root. Polynomial divison produces a quadratric equation with solutions $5(2 - sqrt{3})/2$ and $5(2 + sqrt{3})/2$. Hence we get $a = 5(2 - sqrt{3})/2$ and $b = 2 + sqrt{3}$.
answered Jan 23 at 1:16
Paul FrostPaul Frost
11.6k3934
$endgroup$
The roots can be written as $a, ab, ab^2$. Then $(x-a)(x-ab)(x-ab^2) = x^3 + px^2 + qx +r$. Multiplying the left hand side and comparing yields $p = -a(1+b+b^2), q = a^2b(1+b+b^2), r = -a^3b^3$. This shows that $q^3 = p^3r$.
Since $r = -a^3b^3$, we get $ab = -sqrt[3]{r}$ which is one of the three roots. Thus, if $q^3 = p^3r$, then it is worth to check whether $-sqrt[3]{r}$ is a root.
Note, however, that $q^3 = p^3r$ is not sufficient in order that the roots have the above form. Take for example $q = r = 0$ and $p ne 0$. Then the polynomial has a zero of order two at $0$ and a zero of order one at $-p$.
For the special equation we have $(100/8)^3 125/8 = (5^2 2^2/2^3)^3 5^3/2^3 = 5^9/2^6 =5^9 2^3/2^9 = (5^3 2/2^3)^3 = (250/8)^3$.
This means that we should consider $-sqrt[3]{r} = 5/2$. In fact, this is a root. Polynomial divison produces a quadratric equation with solutions $5(2 - sqrt{3})/2$ and $5(2 + sqrt{3})/2$. Hence we get $a = 5(2 - sqrt{3})/2$ and $b = 2 + sqrt{3}$.
answered Jan 23 at 1:16
Paul FrostPaul Frost
11.6k3934
edited Jan 23 at 15:15
answered Jan 23 at 1:16
Paul FrostPaul Frost
11.6k3934
answered Jan 23 at 1:16
Paul FrostPaul Frost
11.6k3934
answered Jan 23 at 1:16
Paul FrostPaul Frost
11.6k3934
11.6k3934
add a comment |
add a comment |
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Hint: either factor the equation or use Vieta’s formulas.
$endgroup$
– amd
Jan 23 at 0:03