polynomial, find the sum of the inverse roots of this equation.Finding the sum of non-real roots of a polynomial.find a polynomial whose roots are inverse of squares of roots of $x^3+px+q$Find the sum of real roots of a biquadratic equationMost efficient way to find polynomial rootsPolynomial with real rootsFind all the roots of this polynomialHelp with sum and product of roots.Quadratic equation with integral coefficients but no roots are givenFinding the sum of squares of roots of a quartic polynomial.Using binomial coefficients to find sum of roots of a polynomial.
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polynomial, find the sum of the inverse roots of this equation.
Finding the sum of non-real roots of a polynomial.find a polynomial whose roots are inverse of squares of roots of $x^3+px+q$Find the sum of real roots of a biquadratic equationMost efficient way to find polynomial rootsPolynomial with real rootsFind all the roots of this polynomialHelp with sum and product of roots.Quadratic equation with integral coefficients but no roots are givenFinding the sum of squares of roots of a quartic polynomial.Using binomial coefficients to find sum of roots of a polynomial.
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Calculate the sum of the inverse roots of the equation $x^3–7x^2+4x–1=0$.
My development was:
sum = -7
product = 1
Thus, I believe that to find the inverse roots one only has to share the sum with the product, ie (-7/1) = 7, however, the answer given in the question is 4.
what do you think? do you have any formula?
polynomials
edited 7 hours ago
Jean Marie
34.2k4 gold badges26 silver badges60 bronze badges
asked 8 hours ago
funfun funfun
284 bronze badges
$endgroup$
add a comment |
$begingroup$
Calculate the sum of the inverse roots of the equation $x^3–7x^2+4x–1=0$.
My development was:
sum = -7
product = 1
Thus, I believe that to find the inverse roots one only has to share the sum with the product, ie (-7/1) = 7, however, the answer given in the question is 4.
what do you think? do you have any formula?
polynomials
edited 7 hours ago
Jean Marie
34.2k4 gold badges26 silver badges60 bronze badges
asked 8 hours ago
funfun funfun
284 bronze badges
$endgroup$
add a comment |
$begingroup$
Calculate the sum of the inverse roots of the equation $x^3–7x^2+4x–1=0$.
My development was:
sum = -7
product = 1
Thus, I believe that to find the inverse roots one only has to share the sum with the product, ie (-7/1) = 7, however, the answer given in the question is 4.
what do you think? do you have any formula?
polynomials
edited 7 hours ago
Jean Marie
34.2k4 gold badges26 silver badges60 bronze badges
asked 8 hours ago
funfun funfun
284 bronze badges
$endgroup$
Calculate the sum of the inverse roots of the equation $x^3–7x^2+4x–1=0$.
My development was:
sum = -7
product = 1
Thus, I believe that to find the inverse roots one only has to share the sum with the product, ie (-7/1) = 7, however, the answer given in the question is 4.
what do you think? do you have any formula?
polynomials
polynomials
edited 7 hours ago
Jean Marie
34.2k4 gold badges26 silver badges60 bronze badges
asked 8 hours ago
funfun funfun
284 bronze badges
edited 7 hours ago
Jean Marie
34.2k4 gold badges26 silver badges60 bronze badges
asked 8 hours ago
funfun funfun
284 bronze badges
edited 7 hours ago
Jean Marie
34.2k4 gold badges26 silver badges60 bronze badges
edited 7 hours ago
Jean Marie
34.2k4 gold badges26 silver badges60 bronze badges
edited 7 hours ago
Jean Marie
34.2k4 gold badges26 silver badges60 bronze badges
34.2k4 gold badges26 silver badges60 bronze badges
asked 8 hours ago
funfun funfun
284 bronze badges
asked 8 hours ago
funfun funfun
284 bronze badges
asked 8 hours ago
funfun funfun
284 bronze badges
284 bronze badges
add a comment |
add a comment |
2 Answers
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$begingroup$
If we reverse the coefficients of a polynomial with roots $r_i$, we get a polynomial with roots $1/r_i$. In this case, the reversal is $-x^3+4x^2-7x+1=0$. By Viète's formulas, we see that the sum of reciprocals of roots of the original polynomial is $-frac4-1=4$.
answered 8 hours ago
Parcly TaxelParcly Taxel
49.5k13 gold badges79 silver badges117 bronze badges
$endgroup$
add a comment |
$begingroup$
Let the roots be $x_1,x_2,x_3$, all non-zero. Then you need$$frac1x_1+frac1x_2+frac1x_3=fracx_1x_2+x_2x_3+x_3x_1x_1x_2x_3=frac-c/ad/a=-frac cd$$where $x^3-7x^2+4x-1=ax^3+bx^2+cx+d$.
answered 8 hours ago
Shubham JohriShubham Johri
6,8489 silver badges18 bronze badges
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
If we reverse the coefficients of a polynomial with roots $r_i$, we get a polynomial with roots $1/r_i$. In this case, the reversal is $-x^3+4x^2-7x+1=0$. By Viète's formulas, we see that the sum of reciprocals of roots of the original polynomial is $-frac4-1=4$.
answered 8 hours ago
Parcly TaxelParcly Taxel
49.5k13 gold badges79 silver badges117 bronze badges
$endgroup$
add a comment |
$begingroup$
If we reverse the coefficients of a polynomial with roots $r_i$, we get a polynomial with roots $1/r_i$. In this case, the reversal is $-x^3+4x^2-7x+1=0$. By Viète's formulas, we see that the sum of reciprocals of roots of the original polynomial is $-frac4-1=4$.
answered 8 hours ago
Parcly TaxelParcly Taxel
49.5k13 gold badges79 silver badges117 bronze badges
$endgroup$
add a comment |
$begingroup$
If we reverse the coefficients of a polynomial with roots $r_i$, we get a polynomial with roots $1/r_i$. In this case, the reversal is $-x^3+4x^2-7x+1=0$. By Viète's formulas, we see that the sum of reciprocals of roots of the original polynomial is $-frac4-1=4$.
answered 8 hours ago
Parcly TaxelParcly Taxel
49.5k13 gold badges79 silver badges117 bronze badges
$endgroup$
If we reverse the coefficients of a polynomial with roots $r_i$, we get a polynomial with roots $1/r_i$. In this case, the reversal is $-x^3+4x^2-7x+1=0$. By Viète's formulas, we see that the sum of reciprocals of roots of the original polynomial is $-frac4-1=4$.
answered 8 hours ago
Parcly TaxelParcly Taxel
49.5k13 gold badges79 silver badges117 bronze badges
answered 8 hours ago
Parcly TaxelParcly Taxel
49.5k13 gold badges79 silver badges117 bronze badges
answered 8 hours ago
Parcly TaxelParcly Taxel
49.5k13 gold badges79 silver badges117 bronze badges
answered 8 hours ago
Parcly TaxelParcly Taxel
49.5k13 gold badges79 silver badges117 bronze badges
49.5k13 gold badges79 silver badges117 bronze badges
add a comment |
add a comment |
$begingroup$
Let the roots be $x_1,x_2,x_3$, all non-zero. Then you need$$frac1x_1+frac1x_2+frac1x_3=fracx_1x_2+x_2x_3+x_3x_1x_1x_2x_3=frac-c/ad/a=-frac cd$$where $x^3-7x^2+4x-1=ax^3+bx^2+cx+d$.
answered 8 hours ago
Shubham JohriShubham Johri
6,8489 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
Let the roots be $x_1,x_2,x_3$, all non-zero. Then you need$$frac1x_1+frac1x_2+frac1x_3=fracx_1x_2+x_2x_3+x_3x_1x_1x_2x_3=frac-c/ad/a=-frac cd$$where $x^3-7x^2+4x-1=ax^3+bx^2+cx+d$.
answered 8 hours ago
Shubham JohriShubham Johri
6,8489 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
Let the roots be $x_1,x_2,x_3$, all non-zero. Then you need$$frac1x_1+frac1x_2+frac1x_3=fracx_1x_2+x_2x_3+x_3x_1x_1x_2x_3=frac-c/ad/a=-frac cd$$where $x^3-7x^2+4x-1=ax^3+bx^2+cx+d$.
answered 8 hours ago
Shubham JohriShubham Johri
6,8489 silver badges18 bronze badges
$endgroup$
Let the roots be $x_1,x_2,x_3$, all non-zero. Then you need$$frac1x_1+frac1x_2+frac1x_3=fracx_1x_2+x_2x_3+x_3x_1x_1x_2x_3=frac-c/ad/a=-frac cd$$where $x^3-7x^2+4x-1=ax^3+bx^2+cx+d$.
answered 8 hours ago
Shubham JohriShubham Johri
6,8489 silver badges18 bronze badges
answered 8 hours ago
Shubham JohriShubham Johri
6,8489 silver badges18 bronze badges
answered 8 hours ago
Shubham JohriShubham Johri
6,8489 silver badges18 bronze badges
answered 8 hours ago
Shubham JohriShubham Johri
6,8489 silver badges18 bronze badges
6,8489 silver badges18 bronze badges
add a comment |
add a comment |
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