Finding sum of power $5$ in polynomialSolve for $x$, $3sqrt{x+13} = x+9$no. of solution of the equation...
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Finding sum of power $5$ in polynomial
Solve for $x$, $3sqrt{x+13} = x+9$no. of solution of the equation $[x]^2+a[x]+b = 0$ issum of all non real roots of the equation in a bi-quadratic equationreal values of $a$ for which the range of function $ f(x) = frac{x-1}{1-x^2-a}$ does not contain value from $left[-1,1right]$Solve the equation $18x^2-18x sqrt{x}-17x-8 sqrt{x}-2=0$Find the value of $sum frac{1}{beta+gamma}$.product of terms taken $3$ at a time in polynomial expressionProduct of roots of a polynomial decreased from oneHow to solve first-degree equation with two or more numbers to the right of the $=$?Coefficient of $x^{n-3}$ in $prod^{n}_{k=1}(x-k)$
$begingroup$
Consider the function $f(x)=x^4-2x^3-x^2+17x+6$ and another function $g(x)=x^5-3x^4+22x^2-18x-2$ and $f(x)=0$ has a roots $x_{1},x_{2},x_{3},x_{4}.$ Then $sum^{4}_{k=1}g(x_{k})$ is
what i try
$x^4-2x^3-x^2+17x+6=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})$
$displaystyle sum x_{1}=2;;, sum x_{1}x_{2}=-1;;,sum x_{1}x_{2}x_{3}=-17$ and $displaystyle prod x_{1}=6$
$displaystyle sum^{4}_{k=1}g(x_{k})=sum^{4}_{k=1}(x_{k})^5-3sum^{4}_{k=1}(x_{k})^4+22sum^{4}_{k=1}(x_{k})^2-18sum^{4}_{k=1}(x_{k})-2sum^{4}_{k=1}1$
$displaystyle sum^{4}_{k=1}g(x_{k}) = sum^{4}_{k=1}(x_{k})^5-3sum^{4}_{k=1}(x_{k})^4+22(4+2)-18(2)-2(4)$
How do i solve it Help me please
algebra-precalculus
asked Mar 13 at 7:49
jackyjacky
1,235815
$endgroup$
add a comment |
$begingroup$
Consider the function $f(x)=x^4-2x^3-x^2+17x+6$ and another function $g(x)=x^5-3x^4+22x^2-18x-2$ and $f(x)=0$ has a roots $x_{1},x_{2},x_{3},x_{4}.$ Then $sum^{4}_{k=1}g(x_{k})$ is
what i try
$x^4-2x^3-x^2+17x+6=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})$
$displaystyle sum x_{1}=2;;, sum x_{1}x_{2}=-1;;,sum x_{1}x_{2}x_{3}=-17$ and $displaystyle prod x_{1}=6$
$displaystyle sum^{4}_{k=1}g(x_{k})=sum^{4}_{k=1}(x_{k})^5-3sum^{4}_{k=1}(x_{k})^4+22sum^{4}_{k=1}(x_{k})^2-18sum^{4}_{k=1}(x_{k})-2sum^{4}_{k=1}1$
$displaystyle sum^{4}_{k=1}g(x_{k}) = sum^{4}_{k=1}(x_{k})^5-3sum^{4}_{k=1}(x_{k})^4+22(4+2)-18(2)-2(4)$
How do i solve it Help me please
algebra-precalculus
asked Mar 13 at 7:49
jackyjacky
1,235815
$endgroup$
1
$begingroup$
I think you are already on the right track. First, please adjust your notation in the third row. I think you mean $sum x_k=2$ and similarly for the other equations. In the last equation, start at the back. You already know the $sum x_k$ term. For the $sum x_k^2$ term you can use $(sum x_k)^2=sum x_k^2 + 2 sum_{k neq l}x_kx_l$. Similarly but with more algegra required you can compute the remaining terms.
$endgroup$
– quarague
Mar 13 at 8:03
1
$begingroup$
Your task is simplified somewhat if you also use the fact that $$g(x)-(x-1)f(x)=-x^3+4x^2-7x+4.$$ That way you don't have to go higher than $sum_{k=1}^4x_k^3$. Anyway, you may want to use Newton's identities, but the sum of third powers in terms of the elementary symmetric polynomials may be simpler to work out by hand.
$endgroup$
– Jyrki Lahtonen
Mar 13 at 8:07
$begingroup$
@Jyrki Lahtonen did not know How to proceed further, please explain
$endgroup$
– jacky
Mar 13 at 8:20
add a comment |
$begingroup$
Consider the function $f(x)=x^4-2x^3-x^2+17x+6$ and another function $g(x)=x^5-3x^4+22x^2-18x-2$ and $f(x)=0$ has a roots $x_{1},x_{2},x_{3},x_{4}.$ Then $sum^{4}_{k=1}g(x_{k})$ is
what i try
$x^4-2x^3-x^2+17x+6=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})$
$displaystyle sum x_{1}=2;;, sum x_{1}x_{2}=-1;;,sum x_{1}x_{2}x_{3}=-17$ and $displaystyle prod x_{1}=6$
$displaystyle sum^{4}_{k=1}g(x_{k})=sum^{4}_{k=1}(x_{k})^5-3sum^{4}_{k=1}(x_{k})^4+22sum^{4}_{k=1}(x_{k})^2-18sum^{4}_{k=1}(x_{k})-2sum^{4}_{k=1}1$
$displaystyle sum^{4}_{k=1}g(x_{k}) = sum^{4}_{k=1}(x_{k})^5-3sum^{4}_{k=1}(x_{k})^4+22(4+2)-18(2)-2(4)$
How do i solve it Help me please
algebra-precalculus
asked Mar 13 at 7:49
jackyjacky
1,235815
$endgroup$
Consider the function $f(x)=x^4-2x^3-x^2+17x+6$ and another function $g(x)=x^5-3x^4+22x^2-18x-2$ and $f(x)=0$ has a roots $x_{1},x_{2},x_{3},x_{4}.$ Then $sum^{4}_{k=1}g(x_{k})$ is
what i try
$x^4-2x^3-x^2+17x+6=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})$
$displaystyle sum x_{1}=2;;, sum x_{1}x_{2}=-1;;,sum x_{1}x_{2}x_{3}=-17$ and $displaystyle prod x_{1}=6$
$displaystyle sum^{4}_{k=1}g(x_{k})=sum^{4}_{k=1}(x_{k})^5-3sum^{4}_{k=1}(x_{k})^4+22sum^{4}_{k=1}(x_{k})^2-18sum^{4}_{k=1}(x_{k})-2sum^{4}_{k=1}1$
$displaystyle sum^{4}_{k=1}g(x_{k}) = sum^{4}_{k=1}(x_{k})^5-3sum^{4}_{k=1}(x_{k})^4+22(4+2)-18(2)-2(4)$
How do i solve it Help me please
algebra-precalculus
algebra-precalculus
asked Mar 13 at 7:49
jackyjacky
1,235815
asked Mar 13 at 7:49
jackyjacky
1,235815
edited Mar 13 at 8:15
jacky
asked Mar 13 at 7:49
jackyjacky
1,235815
asked Mar 13 at 7:49
jackyjacky
1,235815
asked Mar 13 at 7:49
jackyjacky
1,235815
1,235815
1
$begingroup$
I think you are already on the right track. First, please adjust your notation in the third row. I think you mean $sum x_k=2$ and similarly for the other equations. In the last equation, start at the back. You already know the $sum x_k$ term. For the $sum x_k^2$ term you can use $(sum x_k)^2=sum x_k^2 + 2 sum_{k neq l}x_kx_l$. Similarly but with more algegra required you can compute the remaining terms.
$endgroup$
– quarague
Mar 13 at 8:03
1
$begingroup$
Your task is simplified somewhat if you also use the fact that $$g(x)-(x-1)f(x)=-x^3+4x^2-7x+4.$$ That way you don't have to go higher than $sum_{k=1}^4x_k^3$. Anyway, you may want to use Newton's identities, but the sum of third powers in terms of the elementary symmetric polynomials may be simpler to work out by hand.
$endgroup$
– Jyrki Lahtonen
Mar 13 at 8:07
$begingroup$
@Jyrki Lahtonen did not know How to proceed further, please explain
$endgroup$
– jacky
Mar 13 at 8:20
add a comment |
1
$begingroup$
I think you are already on the right track. First, please adjust your notation in the third row. I think you mean $sum x_k=2$ and similarly for the other equations. In the last equation, start at the back. You already know the $sum x_k$ term. For the $sum x_k^2$ term you can use $(sum x_k)^2=sum x_k^2 + 2 sum_{k neq l}x_kx_l$. Similarly but with more algegra required you can compute the remaining terms.
$endgroup$
– quarague
Mar 13 at 8:03
1
$begingroup$
Your task is simplified somewhat if you also use the fact that $$g(x)-(x-1)f(x)=-x^3+4x^2-7x+4.$$ That way you don't have to go higher than $sum_{k=1}^4x_k^3$. Anyway, you may want to use Newton's identities, but the sum of third powers in terms of the elementary symmetric polynomials may be simpler to work out by hand.
$endgroup$
– Jyrki Lahtonen
Mar 13 at 8:07
$begingroup$
@Jyrki Lahtonen did not know How to proceed further, please explain
$endgroup$
– jacky
Mar 13 at 8:20
1
1
$begingroup$
I think you are already on the right track. First, please adjust your notation in the third row. I think you mean $sum x_k=2$ and similarly for the other equations. In the last equation, start at the back. You already know the $sum x_k$ term. For the $sum x_k^2$ term you can use $(sum x_k)^2=sum x_k^2 + 2 sum_{k neq l}x_kx_l$. Similarly but with more algegra required you can compute the remaining terms.
$endgroup$
– quarague
Mar 13 at 8:03
$begingroup$
I think you are already on the right track. First, please adjust your notation in the third row. I think you mean $sum x_k=2$ and similarly for the other equations. In the last equation, start at the back. You already know the $sum x_k$ term. For the $sum x_k^2$ term you can use $(sum x_k)^2=sum x_k^2 + 2 sum_{k neq l}x_kx_l$. Similarly but with more algegra required you can compute the remaining terms.
$endgroup$
– quarague
Mar 13 at 8:03
1
1
$begingroup$
Your task is simplified somewhat if you also use the fact that $$g(x)-(x-1)f(x)=-x^3+4x^2-7x+4.$$ That way you don't have to go higher than $sum_{k=1}^4x_k^3$. Anyway, you may want to use Newton's identities, but the sum of third powers in terms of the elementary symmetric polynomials may be simpler to work out by hand.
$endgroup$
– Jyrki Lahtonen
Mar 13 at 8:07
$begingroup$
Your task is simplified somewhat if you also use the fact that $$g(x)-(x-1)f(x)=-x^3+4x^2-7x+4.$$ That way you don't have to go higher than $sum_{k=1}^4x_k^3$. Anyway, you may want to use Newton's identities, but the sum of third powers in terms of the elementary symmetric polynomials may be simpler to work out by hand.
$endgroup$
– Jyrki Lahtonen
Mar 13 at 8:07
$begingroup$
@Jyrki Lahtonen did not know How to proceed further, please explain
$endgroup$
– jacky
Mar 13 at 8:20
$begingroup$
@Jyrki Lahtonen did not know How to proceed further, please explain
$endgroup$
– jacky
Mar 13 at 8:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
HINT...Writing $$S_n=sum_{i=1}^4x^n_i$$
You have $$S_0=4, S_1=2, S_2=6$$ And also $$S_{-1}=frac{Sigmaalphabetagamma}{alphabetagammadelta}=-frac{17}{6}$$
Then, you can obtain $S_3$ from the equation itself because $$S_3-2S_2-S_1+17S_0+6S_{-1}=0$$
and then obtain $S_4$ from $$S_4-2S_3-S_2+17S_1+6S_0=0$$
and then obtain $S_5$ from $$S_5-2S_4-S_3+17S_2+6S_1=0$$
Alternatively, following Jyrki Lahtonen's suggestion, you sum the identity he gives so that $$sum g(x_i)-0=-S_3+4S_2-7S_1+4S_0$$
answered Mar 14 at 9:44
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
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$begingroup$
HINT...Writing $$S_n=sum_{i=1}^4x^n_i$$
You have $$S_0=4, S_1=2, S_2=6$$ And also $$S_{-1}=frac{Sigmaalphabetagamma}{alphabetagammadelta}=-frac{17}{6}$$
Then, you can obtain $S_3$ from the equation itself because $$S_3-2S_2-S_1+17S_0+6S_{-1}=0$$
and then obtain $S_4$ from $$S_4-2S_3-S_2+17S_1+6S_0=0$$
and then obtain $S_5$ from $$S_5-2S_4-S_3+17S_2+6S_1=0$$
Alternatively, following Jyrki Lahtonen's suggestion, you sum the identity he gives so that $$sum g(x_i)-0=-S_3+4S_2-7S_1+4S_0$$
answered Mar 14 at 9:44
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
$begingroup$
HINT...Writing $$S_n=sum_{i=1}^4x^n_i$$
You have $$S_0=4, S_1=2, S_2=6$$ And also $$S_{-1}=frac{Sigmaalphabetagamma}{alphabetagammadelta}=-frac{17}{6}$$
Then, you can obtain $S_3$ from the equation itself because $$S_3-2S_2-S_1+17S_0+6S_{-1}=0$$
and then obtain $S_4$ from $$S_4-2S_3-S_2+17S_1+6S_0=0$$
and then obtain $S_5$ from $$S_5-2S_4-S_3+17S_2+6S_1=0$$
Alternatively, following Jyrki Lahtonen's suggestion, you sum the identity he gives so that $$sum g(x_i)-0=-S_3+4S_2-7S_1+4S_0$$
answered Mar 14 at 9:44
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
$begingroup$
HINT...Writing $$S_n=sum_{i=1}^4x^n_i$$
You have $$S_0=4, S_1=2, S_2=6$$ And also $$S_{-1}=frac{Sigmaalphabetagamma}{alphabetagammadelta}=-frac{17}{6}$$
Then, you can obtain $S_3$ from the equation itself because $$S_3-2S_2-S_1+17S_0+6S_{-1}=0$$
and then obtain $S_4$ from $$S_4-2S_3-S_2+17S_1+6S_0=0$$
and then obtain $S_5$ from $$S_5-2S_4-S_3+17S_2+6S_1=0$$
Alternatively, following Jyrki Lahtonen's suggestion, you sum the identity he gives so that $$sum g(x_i)-0=-S_3+4S_2-7S_1+4S_0$$
answered Mar 14 at 9:44
David QuinnDavid Quinn
24.1k21141
$endgroup$
HINT...Writing $$S_n=sum_{i=1}^4x^n_i$$
You have $$S_0=4, S_1=2, S_2=6$$ And also $$S_{-1}=frac{Sigmaalphabetagamma}{alphabetagammadelta}=-frac{17}{6}$$
Then, you can obtain $S_3$ from the equation itself because $$S_3-2S_2-S_1+17S_0+6S_{-1}=0$$
and then obtain $S_4$ from $$S_4-2S_3-S_2+17S_1+6S_0=0$$
and then obtain $S_5$ from $$S_5-2S_4-S_3+17S_2+6S_1=0$$
Alternatively, following Jyrki Lahtonen's suggestion, you sum the identity he gives so that $$sum g(x_i)-0=-S_3+4S_2-7S_1+4S_0$$
answered Mar 14 at 9:44
David QuinnDavid Quinn
24.1k21141
answered Mar 14 at 9:44
David QuinnDavid Quinn
24.1k21141
answered Mar 14 at 9:44
David QuinnDavid Quinn
24.1k21141
answered Mar 14 at 9:44
David QuinnDavid Quinn
24.1k21141
24.1k21141
add a comment |
add a comment |
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$begingroup$
I think you are already on the right track. First, please adjust your notation in the third row. I think you mean $sum x_k=2$ and similarly for the other equations. In the last equation, start at the back. You already know the $sum x_k$ term. For the $sum x_k^2$ term you can use $(sum x_k)^2=sum x_k^2 + 2 sum_{k neq l}x_kx_l$. Similarly but with more algegra required you can compute the remaining terms.
$endgroup$
– quarague
Mar 13 at 8:03
1
$begingroup$
Your task is simplified somewhat if you also use the fact that $$g(x)-(x-1)f(x)=-x^3+4x^2-7x+4.$$ That way you don't have to go higher than $sum_{k=1}^4x_k^3$. Anyway, you may want to use Newton's identities, but the sum of third powers in terms of the elementary symmetric polynomials may be simpler to work out by hand.
$endgroup$
– Jyrki Lahtonen
Mar 13 at 8:07
$begingroup$
@Jyrki Lahtonen did not know How to proceed further, please explain
$endgroup$
– jacky
Mar 13 at 8:20