need help on polynomial (1 Viewer)

[noobking]

: given that 3+2^0.5 is the root of the equation : x^3-11x^2+kx+l=0 . find the other roots of the equation

 

[airie]

: given that 3+2^0.5 is the root of the equation : x^3-11x^2+kx+l=0 . find the other roots of the equation

Use Vieta's theorem aka relationship between sum and product of roots and the coefficients. So let the other roots be a and b, you get:
- a + b + (3+2^0.5) = 11
- ab + b(3+2^0.5) + a(3+2^0.5) = k
- ab(3+2^0.5) = -l

Then solve simultaneously.

 

[jyu]

Use Vieta's theorem aka relationship between sum and product of roots and the coefficients. So let the other roots be a and b, you get:
- a + b + (3+2^0.5) = 11
- ab + b(3+2^0.5) + a(3+2^0.5) = k
- ab(3+2^0.5) = -l

Then solve simultaneously.

How do you solve these equations? 3 equations for 4 unknowns or in terms of k and l?

 

[LoneShadow]

You have 2 unknowns: k and l. So you need 2 equations in these 2 unknowns.

call f(x) = x^3-11x^2+kx+l

The first equation comes from substituting x = 3+Sqrt[2]:
f(3+Sqrt[2]) = -166 - 95*Sqrt[2] - (3+Sqrt[2])*k + l = 0 (since 3+Sqrt[2] is a root]

This gives: l - (3+Sqrt[2])*k = 166 + 95*Sqrt[2] ----(1)

The second equation comes from long division of f(x) by x-(3+Sqrt[2]):

f(x) = [x-(3+Sqrt[2])]*[x^2 +(Sqrt[2]-8)*x+(k-22-5*Sqrt[2])] + [l - 76 -37*Sqrt[2] + (3+Sqrt[2])*k]

which is in the form of: f(x) = divisor*quotient + remainder. Since x-(3+Sqrt[2]) is a root of f(x), it divides f(x). Hence remainder = 0.

i.e. l - 76 -37*Sqrt[2] + (3+Sqrt[2])*k = 0 which gives:
l + (3+Sqrt[2])*k = 76-37*Sqrt[2] ----(2)

Solve (1) and (2) simultaneously to get values of k and l. Then substitude this into "quotient" [i.e. x^2 +(Sqrt[2]-8)*x+(k-22-5*Sqrt[2])] so f(x) looks like this:
f(x) = (x-a)*(x^2+cx+d). I'm sure you can factorize the quadratic part to get complete factorization of f(x) and hence the roots.

 

[Wackedupwacko]

since all coefficients are real integers (im assuming k and l are integers) , 3- root2 must also be a root.

3+root2+3-root2 + a = 11

therefore a = 5

 

[jyu]

You have 2 unknowns: k and l. So you need 2 equations in these 2 unknowns.

call f(x) = x^3-11x^2+kx+l

The first equation comes from substituting x = 3+Sqrt[2]:
f(3+Sqrt[2]) = -166 - 95*Sqrt[2] - (3+Sqrt[2])*k + l = 0 (since 3+Sqrt[2] is a root]

This gives: l - (3+Sqrt[2])*k = 166 + 95*Sqrt[2] ----(1)

The second equation comes from long division of f(x) by x-(3+Sqrt[2]):

f(x) = [x-(3+Sqrt[2])]*[x^2 +(Sqrt[2]-8)*x+(k-22-5*Sqrt[2])] + [l - 76 -37*Sqrt[2] + (3+Sqrt[2])*k]

which is in the form of: f(x) = divisor*quotient + remainder. Since x-(3+Sqrt[2]) is a root of f(x), it divides f(x). Hence remainder = 0.

i.e. l - 76 -37*Sqrt[2] + (3+Sqrt[2])*k = 0 which gives:
l + (3+Sqrt[2])*k = 76-37*Sqrt[2] ----(2)

Solve (1) and (2) simultaneously to get values of k and l. Then substitude this into "quotient" [i.e. x^2 +(Sqrt[2]-8)*x+(k-22-5*Sqrt[2])] so f(x) looks like this:
f(x) = (x-a)*(x^2+cx+d). I'm sure you can factorize the quadratic part to get complete factorization of f(x) and hence the roots.

Basically you used the same information to obtain the two equations in k and l, and they should be the same. You don't end up with two different equations.

 

[jyu]

since all coefficients are real integers (im assuming k and l are integers) , 3- root2 must also be a root.

3+root2+3-root2 + a = 11

therefore a = 5

The original question did not specify that they are integers! Your assumption may not be the intention of the question.

 

[Wackedupwacko]

x.x well true but... usualyl they specify what k and l are... i mean for all you know they could be complex =P but in all likely hood itll have somewhere where k and l are integers x.x

if not then just simultaneously solve the eqns in airies post treating k and l a s constants (so you will end up with roots that have k and l in it)

 

[airie]

How do you solve these equations? 3 equations for 4 unknowns or in terms of k and l?

You solve for a and b in terms of k and l, since you're not told what are the values of k and l.