Mixing
I need help with a competition math problem please. [duplicate]
$begingroup$
This question already has an answer here:
Solving a polynomial with grouping
1 answer
I am a competition math student, and I was doing a practice test when I found this problem:
"The fourth degree polynomial equation $x^4 - 7x^3 + 4x^2 + 7x - 4 = 0$ has four real roots a, b, c and d. What is the value of the sum $1/a + 1/b + 1/c + 1/d$? Express your answer as a common fraction."
First of all, I don't think this polynomial can be factored, so I don't know how to find the reciprocals of the roots. I need help and any help would be appreciated! Note: When they say "Express your answer as a common fraction.", it means that the answer IS a fraction. I tried asking this question already, but I worded it differently, and they didn't get a fraction so I knew it was wrong. This is the link to my other question: Link
algebra-precalculus polynomials
asked Jan 30 at 0:19
Spencer1O1Spencer1O1
125
$endgroup$
marked as duplicate by jmerry, Andrew, Robert Soupe, Lord Shark the Unknown, Leucippus Jan 30 at 8:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Solving a polynomial with grouping
1 answer
I am a competition math student, and I was doing a practice test when I found this problem:
"The fourth degree polynomial equation $x^4 - 7x^3 + 4x^2 + 7x - 4 = 0$ has four real roots a, b, c and d. What is the value of the sum $1/a + 1/b + 1/c + 1/d$? Express your answer as a common fraction."
First of all, I don't think this polynomial can be factored, so I don't know how to find the reciprocals of the roots. I need help and any help would be appreciated! Note: When they say "Express your answer as a common fraction.", it means that the answer IS a fraction. I tried asking this question already, but I worded it differently, and they didn't get a fraction so I knew it was wrong. This is the link to my other question: Link
algebra-precalculus polynomials
asked Jan 30 at 0:19
Spencer1O1Spencer1O1
125
$endgroup$
marked as duplicate by jmerry, Andrew, Robert Soupe, Lord Shark the Unknown, Leucippus Jan 30 at 8:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
You may find Vieta's formulas useful.
$endgroup$
– nathan.j.mcdougall
Jan 30 at 0:23
$begingroup$
Make the substitution $y=1/x$ and consider the quartic in $y$.
$endgroup$
– Donald Splutterwit
Jan 30 at 0:23
$begingroup$
See here
$endgroup$
– J. W. Tanner
Jan 30 at 0:34
$begingroup$
@J.W.Tanner It isn't a duplicate. Read my edit.
$endgroup$
– Spencer1O1
Jan 30 at 0:42
$begingroup$
The answer there has been corrected now
$endgroup$
– J. W. Tanner
Jan 30 at 17:40
add a comment |
$begingroup$
This question already has an answer here:
Solving a polynomial with grouping
1 answer
I am a competition math student, and I was doing a practice test when I found this problem:
"The fourth degree polynomial equation $x^4 - 7x^3 + 4x^2 + 7x - 4 = 0$ has four real roots a, b, c and d. What is the value of the sum $1/a + 1/b + 1/c + 1/d$? Express your answer as a common fraction."
First of all, I don't think this polynomial can be factored, so I don't know how to find the reciprocals of the roots. I need help and any help would be appreciated! Note: When they say "Express your answer as a common fraction.", it means that the answer IS a fraction. I tried asking this question already, but I worded it differently, and they didn't get a fraction so I knew it was wrong. This is the link to my other question: Link
algebra-precalculus polynomials
asked Jan 30 at 0:19
Spencer1O1Spencer1O1
125
$endgroup$
This question already has an answer here:
Solving a polynomial with grouping
1 answer
I am a competition math student, and I was doing a practice test when I found this problem:
"The fourth degree polynomial equation $x^4 - 7x^3 + 4x^2 + 7x - 4 = 0$ has four real roots a, b, c and d. What is the value of the sum $1/a + 1/b + 1/c + 1/d$? Express your answer as a common fraction."
First of all, I don't think this polynomial can be factored, so I don't know how to find the reciprocals of the roots. I need help and any help would be appreciated! Note: When they say "Express your answer as a common fraction.", it means that the answer IS a fraction. I tried asking this question already, but I worded it differently, and they didn't get a fraction so I knew it was wrong. This is the link to my other question: Link
This question already has an answer here:
Solving a polynomial with grouping
1 answer
algebra-precalculus polynomials
algebra-precalculus polynomials
asked Jan 30 at 0:19
Spencer1O1Spencer1O1
125
asked Jan 30 at 0:19
Spencer1O1Spencer1O1
125
edited Jan 30 at 0:40
Spencer1O1
asked Jan 30 at 0:19
Spencer1O1Spencer1O1
125
asked Jan 30 at 0:19
Spencer1O1Spencer1O1
125
asked Jan 30 at 0:19
Spencer1O1Spencer1O1
125
125
marked as duplicate by jmerry, Andrew, Robert Soupe, Lord Shark the Unknown, Leucippus Jan 30 at 8:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by jmerry, Andrew, Robert Soupe, Lord Shark the Unknown, Leucippus Jan 30 at 8:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
You may find Vieta's formulas useful.
$endgroup$
– nathan.j.mcdougall
Jan 30 at 0:23
$begingroup$
Make the substitution $y=1/x$ and consider the quartic in $y$.
$endgroup$
– Donald Splutterwit
Jan 30 at 0:23
$begingroup$
See here
$endgroup$
– J. W. Tanner
Jan 30 at 0:34
$begingroup$
@J.W.Tanner It isn't a duplicate. Read my edit.
$endgroup$
– Spencer1O1
Jan 30 at 0:42
$begingroup$
The answer there has been corrected now
$endgroup$
– J. W. Tanner
Jan 30 at 17:40
add a comment |
$begingroup$
You may find Vieta's formulas useful.
$endgroup$
– nathan.j.mcdougall
Jan 30 at 0:23
$begingroup$
Make the substitution $y=1/x$ and consider the quartic in $y$.
$endgroup$
– Donald Splutterwit
Jan 30 at 0:23
$begingroup$
See here
$endgroup$
– J. W. Tanner
Jan 30 at 0:34
$begingroup$
@J.W.Tanner It isn't a duplicate. Read my edit.
$endgroup$
– Spencer1O1
Jan 30 at 0:42
$begingroup$
The answer there has been corrected now
$endgroup$
– J. W. Tanner
Jan 30 at 17:40
$begingroup$
You may find Vieta's formulas useful.
$endgroup$
– nathan.j.mcdougall
Jan 30 at 0:23
$begingroup$
You may find Vieta's formulas useful.
$endgroup$
– nathan.j.mcdougall
Jan 30 at 0:23
$begingroup$
Make the substitution $y=1/x$ and consider the quartic in $y$.
$endgroup$
– Donald Splutterwit
Jan 30 at 0:23
$begingroup$
Make the substitution $y=1/x$ and consider the quartic in $y$.
$endgroup$
– Donald Splutterwit
Jan 30 at 0:23
$begingroup$
See here
$endgroup$
– J. W. Tanner
Jan 30 at 0:34
$begingroup$
See here
$endgroup$
– J. W. Tanner
Jan 30 at 0:34
$begingroup$
@J.W.Tanner It isn't a duplicate. Read my edit.
$endgroup$
– Spencer1O1
Jan 30 at 0:42
$begingroup$
@J.W.Tanner It isn't a duplicate. Read my edit.
$endgroup$
– Spencer1O1
Jan 30 at 0:42
$begingroup$
The answer there has been corrected now
$endgroup$
– J. W. Tanner
Jan 30 at 17:40
$begingroup$
The answer there has been corrected now
$endgroup$
– J. W. Tanner
Jan 30 at 17:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose we have a polynomial $f(x)=a_nx^n+a_{n-1}x^{n-1}+dots +a_0$ with roots $r_1,r_2,dots, r_n$. We claim that the polynomial with roots $frac{1}{r_1},frac{1}{r_2},dots,frac{1}{r_n}$ is simply $g(x)=a_0x^n+a_1x^{n-1}+dots + a_n$. Why? Because if you observe that the roots of the polynomial $f(frac{1}{x})$ are the reciprocals of the roots of $f(x)$, then our desired polynomial is $$begin{align*}g(x)&=x^nf(1/x)\&=x^nleft(frac{a_n}{x^n}+frac{a_{n-1}} {x^{n-1}}+cdots+frac{a_1}{x}+a_0right)\&=a_0x^n+a_1x^{n-1}+cdots+a_n,end{align*}$$
The desired polynomial can be found by simply 'reversing the coefficients' as shown above. Hence the polynomial with roots $frac{1}{a},frac 1b, frac 1c,$ and $frac 1d$ is $$-4x^4+7x^3+4x^2-7x+1$$ Applying Vieta's Formulae to this, we find that the sum of the roots is $-frac{-7}{4}=frac{7}{4}$.
answered Jan 30 at 0:30
user574848user574848
688118
$endgroup$
1
$begingroup$
Thanks! I got 7/4 too because my reasoning was that $x^4 p(1/x) = -4x^4+7x^3+4x^2-7x+1$ and the sum of the roots is $-7/-4=7/4$. I just wasn't 100% sure that you could do this.
$endgroup$
– Spencer1O1
Jan 30 at 0:38
$begingroup$
@Spencer1O1 nice work! In the future, consider tagging your competition math questions with the tag [contest-math]
$endgroup$
– user574848
Jan 30 at 0:40
$begingroup$
Okay! I tried to tag it with [competition-math], but it didn't exist. Thanks for the tip!
$endgroup$
– Spencer1O1
Jan 30 at 0:46
add a comment |
$begingroup$
The trick is to write
$$
frac{1}{a}+
frac{1}{b}+
frac{1}{c}+
frac{1}{d}=frac{bcd+acd+abd+abc}{abcd}
$$
write $f(x)=x^4-7x^3+4x^2+7x-4=(x-a)(x-b)(x-c)(x-d)$ and use Vieta's relations.
For example, by multiplying out we see at once that $abcd=-4$ and $-(bcd+acd+abd+abc)=7.$
answered Jan 30 at 0:26
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose we have a polynomial $f(x)=a_nx^n+a_{n-1}x^{n-1}+dots +a_0$ with roots $r_1,r_2,dots, r_n$. We claim that the polynomial with roots $frac{1}{r_1},frac{1}{r_2},dots,frac{1}{r_n}$ is simply $g(x)=a_0x^n+a_1x^{n-1}+dots + a_n$. Why? Because if you observe that the roots of the polynomial $f(frac{1}{x})$ are the reciprocals of the roots of $f(x)$, then our desired polynomial is $$begin{align*}g(x)&=x^nf(1/x)\&=x^nleft(frac{a_n}{x^n}+frac{a_{n-1}} {x^{n-1}}+cdots+frac{a_1}{x}+a_0right)\&=a_0x^n+a_1x^{n-1}+cdots+a_n,end{align*}$$
The desired polynomial can be found by simply 'reversing the coefficients' as shown above. Hence the polynomial with roots $frac{1}{a},frac 1b, frac 1c,$ and $frac 1d$ is $$-4x^4+7x^3+4x^2-7x+1$$ Applying Vieta's Formulae to this, we find that the sum of the roots is $-frac{-7}{4}=frac{7}{4}$.
answered Jan 30 at 0:30
user574848user574848
688118
$endgroup$
1
$begingroup$
Thanks! I got 7/4 too because my reasoning was that $x^4 p(1/x) = -4x^4+7x^3+4x^2-7x+1$ and the sum of the roots is $-7/-4=7/4$. I just wasn't 100% sure that you could do this.
$endgroup$
– Spencer1O1
Jan 30 at 0:38
$begingroup$
@Spencer1O1 nice work! In the future, consider tagging your competition math questions with the tag [contest-math]
$endgroup$
– user574848
Jan 30 at 0:40
$begingroup$
Okay! I tried to tag it with [competition-math], but it didn't exist. Thanks for the tip!
$endgroup$
– Spencer1O1
Jan 30 at 0:46
add a comment |
$begingroup$
Suppose we have a polynomial $f(x)=a_nx^n+a_{n-1}x^{n-1}+dots +a_0$ with roots $r_1,r_2,dots, r_n$. We claim that the polynomial with roots $frac{1}{r_1},frac{1}{r_2},dots,frac{1}{r_n}$ is simply $g(x)=a_0x^n+a_1x^{n-1}+dots + a_n$. Why? Because if you observe that the roots of the polynomial $f(frac{1}{x})$ are the reciprocals of the roots of $f(x)$, then our desired polynomial is $$begin{align*}g(x)&=x^nf(1/x)\&=x^nleft(frac{a_n}{x^n}+frac{a_{n-1}} {x^{n-1}}+cdots+frac{a_1}{x}+a_0right)\&=a_0x^n+a_1x^{n-1}+cdots+a_n,end{align*}$$
The desired polynomial can be found by simply 'reversing the coefficients' as shown above. Hence the polynomial with roots $frac{1}{a},frac 1b, frac 1c,$ and $frac 1d$ is $$-4x^4+7x^3+4x^2-7x+1$$ Applying Vieta's Formulae to this, we find that the sum of the roots is $-frac{-7}{4}=frac{7}{4}$.
answered Jan 30 at 0:30
user574848user574848
688118
$endgroup$
1
$begingroup$
Thanks! I got 7/4 too because my reasoning was that $x^4 p(1/x) = -4x^4+7x^3+4x^2-7x+1$ and the sum of the roots is $-7/-4=7/4$. I just wasn't 100% sure that you could do this.
$endgroup$
– Spencer1O1
Jan 30 at 0:38
$begingroup$
@Spencer1O1 nice work! In the future, consider tagging your competition math questions with the tag [contest-math]
$endgroup$
– user574848
Jan 30 at 0:40
$begingroup$
Okay! I tried to tag it with [competition-math], but it didn't exist. Thanks for the tip!
$endgroup$
– Spencer1O1
Jan 30 at 0:46
add a comment |
$begingroup$
Suppose we have a polynomial $f(x)=a_nx^n+a_{n-1}x^{n-1}+dots +a_0$ with roots $r_1,r_2,dots, r_n$. We claim that the polynomial with roots $frac{1}{r_1},frac{1}{r_2},dots,frac{1}{r_n}$ is simply $g(x)=a_0x^n+a_1x^{n-1}+dots + a_n$. Why? Because if you observe that the roots of the polynomial $f(frac{1}{x})$ are the reciprocals of the roots of $f(x)$, then our desired polynomial is $$begin{align*}g(x)&=x^nf(1/x)\&=x^nleft(frac{a_n}{x^n}+frac{a_{n-1}} {x^{n-1}}+cdots+frac{a_1}{x}+a_0right)\&=a_0x^n+a_1x^{n-1}+cdots+a_n,end{align*}$$
The desired polynomial can be found by simply 'reversing the coefficients' as shown above. Hence the polynomial with roots $frac{1}{a},frac 1b, frac 1c,$ and $frac 1d$ is $$-4x^4+7x^3+4x^2-7x+1$$ Applying Vieta's Formulae to this, we find that the sum of the roots is $-frac{-7}{4}=frac{7}{4}$.
answered Jan 30 at 0:30
user574848user574848
688118
$endgroup$
Suppose we have a polynomial $f(x)=a_nx^n+a_{n-1}x^{n-1}+dots +a_0$ with roots $r_1,r_2,dots, r_n$. We claim that the polynomial with roots $frac{1}{r_1},frac{1}{r_2},dots,frac{1}{r_n}$ is simply $g(x)=a_0x^n+a_1x^{n-1}+dots + a_n$. Why? Because if you observe that the roots of the polynomial $f(frac{1}{x})$ are the reciprocals of the roots of $f(x)$, then our desired polynomial is $$begin{align*}g(x)&=x^nf(1/x)\&=x^nleft(frac{a_n}{x^n}+frac{a_{n-1}} {x^{n-1}}+cdots+frac{a_1}{x}+a_0right)\&=a_0x^n+a_1x^{n-1}+cdots+a_n,end{align*}$$
The desired polynomial can be found by simply 'reversing the coefficients' as shown above. Hence the polynomial with roots $frac{1}{a},frac 1b, frac 1c,$ and $frac 1d$ is $$-4x^4+7x^3+4x^2-7x+1$$ Applying Vieta's Formulae to this, we find that the sum of the roots is $-frac{-7}{4}=frac{7}{4}$.
answered Jan 30 at 0:30
user574848user574848
688118
edited Jan 30 at 0:38
answered Jan 30 at 0:30
user574848user574848
688118
answered Jan 30 at 0:30
user574848user574848
688118
answered Jan 30 at 0:30
user574848user574848
688118
688118
1
$begingroup$
Thanks! I got 7/4 too because my reasoning was that $x^4 p(1/x) = -4x^4+7x^3+4x^2-7x+1$ and the sum of the roots is $-7/-4=7/4$. I just wasn't 100% sure that you could do this.
$endgroup$
– Spencer1O1
Jan 30 at 0:38
$begingroup$
@Spencer1O1 nice work! In the future, consider tagging your competition math questions with the tag [contest-math]
$endgroup$
– user574848
Jan 30 at 0:40
$begingroup$
Okay! I tried to tag it with [competition-math], but it didn't exist. Thanks for the tip!
$endgroup$
– Spencer1O1
Jan 30 at 0:46
add a comment |
1
$begingroup$
Thanks! I got 7/4 too because my reasoning was that $x^4 p(1/x) = -4x^4+7x^3+4x^2-7x+1$ and the sum of the roots is $-7/-4=7/4$. I just wasn't 100% sure that you could do this.
$endgroup$
– Spencer1O1
Jan 30 at 0:38
$begingroup$
@Spencer1O1 nice work! In the future, consider tagging your competition math questions with the tag [contest-math]
$endgroup$
– user574848
Jan 30 at 0:40
$begingroup$
Okay! I tried to tag it with [competition-math], but it didn't exist. Thanks for the tip!
$endgroup$
– Spencer1O1
Jan 30 at 0:46
1
1
$begingroup$
Thanks! I got 7/4 too because my reasoning was that $x^4 p(1/x) = -4x^4+7x^3+4x^2-7x+1$ and the sum of the roots is $-7/-4=7/4$. I just wasn't 100% sure that you could do this.
$endgroup$
– Spencer1O1
Jan 30 at 0:38
$begingroup$
Thanks! I got 7/4 too because my reasoning was that $x^4 p(1/x) = -4x^4+7x^3+4x^2-7x+1$ and the sum of the roots is $-7/-4=7/4$. I just wasn't 100% sure that you could do this.
$endgroup$
– Spencer1O1
Jan 30 at 0:38
$begingroup$
@Spencer1O1 nice work! In the future, consider tagging your competition math questions with the tag [contest-math]
$endgroup$
– user574848
Jan 30 at 0:40
$begingroup$
@Spencer1O1 nice work! In the future, consider tagging your competition math questions with the tag [contest-math]
$endgroup$
– user574848
Jan 30 at 0:40
$begingroup$
Okay! I tried to tag it with [competition-math], but it didn't exist. Thanks for the tip!
$endgroup$
– Spencer1O1
Jan 30 at 0:46
$begingroup$
Okay! I tried to tag it with [competition-math], but it didn't exist. Thanks for the tip!
$endgroup$
– Spencer1O1
Jan 30 at 0:46
add a comment |
$begingroup$
The trick is to write
$$
frac{1}{a}+
frac{1}{b}+
frac{1}{c}+
frac{1}{d}=frac{bcd+acd+abd+abc}{abcd}
$$
write $f(x)=x^4-7x^3+4x^2+7x-4=(x-a)(x-b)(x-c)(x-d)$ and use Vieta's relations.
For example, by multiplying out we see at once that $abcd=-4$ and $-(bcd+acd+abd+abc)=7.$
answered Jan 30 at 0:26
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
add a comment |
$begingroup$
The trick is to write
$$
frac{1}{a}+
frac{1}{b}+
frac{1}{c}+
frac{1}{d}=frac{bcd+acd+abd+abc}{abcd}
$$
write $f(x)=x^4-7x^3+4x^2+7x-4=(x-a)(x-b)(x-c)(x-d)$ and use Vieta's relations.
For example, by multiplying out we see at once that $abcd=-4$ and $-(bcd+acd+abd+abc)=7.$
answered Jan 30 at 0:26
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
add a comment |
$begingroup$
The trick is to write
$$
frac{1}{a}+
frac{1}{b}+
frac{1}{c}+
frac{1}{d}=frac{bcd+acd+abd+abc}{abcd}
$$
write $f(x)=x^4-7x^3+4x^2+7x-4=(x-a)(x-b)(x-c)(x-d)$ and use Vieta's relations.
For example, by multiplying out we see at once that $abcd=-4$ and $-(bcd+acd+abd+abc)=7.$
answered Jan 30 at 0:26
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
The trick is to write
$$
frac{1}{a}+
frac{1}{b}+
frac{1}{c}+
frac{1}{d}=frac{bcd+acd+abd+abc}{abcd}
$$
write $f(x)=x^4-7x^3+4x^2+7x-4=(x-a)(x-b)(x-c)(x-d)$ and use Vieta's relations.
For example, by multiplying out we see at once that $abcd=-4$ and $-(bcd+acd+abd+abc)=7.$
answered Jan 30 at 0:26
Foobaz JohnFoobaz John
22.9k41552
answered Jan 30 at 0:26
Foobaz JohnFoobaz John
22.9k41552
answered Jan 30 at 0:26
Foobaz JohnFoobaz John
22.9k41552
answered Jan 30 at 0:26
Foobaz JohnFoobaz John
22.9k41552
22.9k41552
add a comment |
add a comment |
$begingroup$
You may find Vieta's formulas useful.
$endgroup$
– nathan.j.mcdougall
Jan 30 at 0:23
$begingroup$
Make the substitution $y=1/x$ and consider the quartic in $y$.
$endgroup$
– Donald Splutterwit
Jan 30 at 0:23
$begingroup$
See here
$endgroup$
– J. W. Tanner
Jan 30 at 0:34
$begingroup$
@J.W.Tanner It isn't a duplicate. Read my edit.
$endgroup$
– Spencer1O1
Jan 30 at 0:42
$begingroup$
The answer there has been corrected now
$endgroup$
– J. W. Tanner
Jan 30 at 17:40