Polynomial query (1 Viewer)

[mathsbrain]

If p(x)=x^4+kx^3+x^2+x+1, what values of k will all roots be integers?
is it the set k inside Z?

 

[cineti970128]

whats the answer?
i ve got +4, -4 or 0

i am not sure though

 

[cineti970128]

roots are integers, let the roots be (a,b,c,d)
since a*b*c*d = 1
hence roots are 1,1,1,1 or -1-1,1,1 or -1,-1,-1,-1
-k = 1+1+1+1, -1+-1+1+1 or -1+-1+-1+-1
hence k = +4,-4 or 0?

 

[cineti970128]

oh wait you gotta to take in fact that not all 3 solutions satisfy the equation
shit... i dunno anyone

 

[HeroicPandas]

If i make a mistake please let me know (been off the comp for a while)

Before u read this - every 'GREATER THAN OR EQUAL TO' sign u see must be changed to a 'GREATER THAN' sign!!!

For REAL ROOTS:



Solve this inequality and there u have it

k > sqrt{2}

and

k< -sqrt{2}

 

[Menomaths]

If i make a mistake please let me know (been off the comp for a while)

Before u read this - every 'GREATER THAN OR EQUAL TO' sign u see must be changed to a 'GREATER THAN' sign!!!

For REAL ROOTS:



Solve this inequality and there u have it

k > sqrt{2}

and

k< -sqrt{2}

The panda strikes back!

 

[Carrotsticks]

If i make a mistake please let me know (been off the comp for a while)

1. The question was 'What values of K does it have integer roots" not 'real'.

2. Your working out doesn't quite work because you're assuming the converse of the condition.


If the roots are real, then the sum squared > 0.

However, if the sum squared > 0, it does NOT necessarily imply that the roots are real.

Counter: k=root(3) yields no real roots, but it is still > root(2).

 

[mathsbrain]

1. The question was 'What values of K does it have integer roots" not 'real'.

2. Your working out doesn't quite work because you're assuming the converse of the condition.


If the roots are real, then the sum squared > 0.

However, if the sum squared > 0, it does NOT necessarily imply that the roots are real.

Counter: k=root(3) yields no real roots, but it is still > root(2).

Carrot can you help?

 

[Trebla]

roots are integers, let the roots be (a,b,c,d)
since a*b*c*d = 1
hence roots are 1,1,1,1 or -1-1,1,1 or -1,-1,-1,-1
-k = 1+1+1+1, -1+-1+1+1 or -1+-1+-1+-1
hence k = +4,-4 or 0?

Pretty sure your method is correct as there are no other integer roots which satisfy this product of roots condition, but you need to take into consideration the other coefficients. Let the polynomial be P(x).

If P(1) = 0 then k = -4
If P(-1) = 0 then k = 2

Since we obtain different values of k then the polynomial cannot have 1 and -1 simultaneously as roots. This allows us to deduce that the roots must either be all 1 or all -1. If they are all -1 then by the sum of roots k should be 4 but this contradicts the value of k we found based on P(-1) = 0. Hence, we are left with k = -4 which satisfies the sum of roots property for the value of k we got earlier. But the sum of roots in pairs doesn't work out to the right coefficient. Hence, there are no solutions for k.

 

[cineti970128]

Pretty sure your method is correct as there are no other integer roots which satisfy this product of roots condition, but you need to take into consideration the other coefficients. Let the polynomial be P(x).

If P(1) = 0 then k = -4
If P(-1) = 0 then k = 2

Since we obtain different values of k then the polynomial cannot have 1 and -1 simultaneously as roots. This allows us to deduce that the roots must either be all 1 or all -1. If they are all -1 then by the sum of roots k should be 4 but this contradicts the value of k we found based on P(-1) = 0. Hence, we are left with k = -4 which satisfies the sum of roots property for the value of k we got earlier. But the sum of roots in pairs doesn't work out to the right coefficient. Hence, there are no solutions for k.

Ummm I did say that it didn't work following on what I had said

However so there is no solution?

 

[HeroicPandas]

1. The question was 'What values of K does it have integer roots" not 'real'.

2. Your working out doesn't quite work because you're assuming the converse of the condition.


If the roots are real, then the sum squared > 0.

However, if the sum squared > 0, it does NOT necessarily imply that the roots are real.

Counter: k=root(3) yields no real roots, but it is still > root(2).

thanks!! i understand now