Polynomials help (1 Viewer)

[planino]

Could someone please help me with this?

If the sum of two roots of
x⁴ + 2x³ - 8x² - 18x - 9 - 0 = is 0, find the roots of the equation.

 

[SpiralFlex]

As the sum of two roots is 0, we will select alpha + beta to be equal to 0.






Notice if we can somehow factorise like this [Something](alpha + beta) we will eliminate two terms.








Similar process to before,





From (2),




Now some information arises,




Form a quadratic with each of these.

 

[Carrotsticks]

Let the four roots be k, -k, A and B.

Sum of roots is A + B + k - k = -2, hence A + B = -2.

Sum of roots in pairs is -k^2 + Ak + Bk - Ak - Bk + AB = -8, hence -k^2 + AB = -8.

Product of roots is -k^2 x AB = -9

Substituting the fact that AB = k^2 - 8, we have -k^2 (k^2 - 8) = -9

Re-arranging this, we get a quartic (hidden as a quadratic): k^4 - 8k^2 - 9 = 0.

Solving this, we have (k^2-9)(k^2+1) = 0 and hence k = plus/minus 3.

Now, substitute this back into the product of roots to acquire -9 x AB = -9 and hence AB = 1.

We now have the simultaneous equations A + B = -2 and AB = 1, which I am sure you can finish off yourself. Trivially the solutions are A=-1 and B=-1, but your school may be strict with it and require you to actually solve a quadratic etc.

 

[planino]

Wow, thank you so much!
Much appreciated

 

[planino]

Thanks, Carrotsticks

 

[superSAIyan2]

Just out of curiosity is it possible for a quarticsuch as the one above to have 4 real roots as well as imaginary roots even though the rule is that the max amount of roots a poly can have is its degree.

The reason i asked is because in carrotsticks' working he got (k^2-9)(k^2+1)=0 -----> so could k = plus/minus i as well? Obviously this won't be asked in 3unit. But is it still a root? However when u sub x=i into the equation you dont get zero.

Can anyone explain this to me? Carrotsticks do you have any suggestions ?
BTW i have not learnt polynomials in 4unit yet so i dont know any associated theorems. I'm just being curious.

 

[RealiseNothing]

However when u sub x=i into the equation you dont get zero.

You do get 0.

 

[Sy123]

You do get 0.

We get -20i

 

[RealiseNothing]

We get -20i

Substitute 'i' into



???

 

[Sy123]

Just out of curiosity is it possible for a quarticsuch as the one above to have 4 real roots as well as imaginary roots even though the rule is that the max amount of roots a poly can have is its degree.

The reason i asked is because in carrotsticks' working he got (k^2-9)(k^2+1)=0 -----> so could k = plus/minus i as well? Obviously this won't be asked in 3unit. But is it still a root? However when u sub x=i into the equation you dont get zero.

Can anyone explain this to me? Carrotsticks do you have any suggestions ?
BTW i have not learnt polynomials in 4unit yet so i dont know any associated theorems. I'm just being curious.

The reason why we do not get zeroes from is when we are solving for k in C in this particular case.
I am guessing it is similar to the reason why some solutions do not work when we are solving for multiple absolute values:
For instance:



When we solve for this we can 4 solutions, however sometimes only 2 of them are correct, and we find if they are correct or not via substitution. Probably NOT because of the same reason, some of the solutions in k just dont work. And I am guessing it is similar logic.

The fundamental theorem of algebra will always hold for x in C.
Though I honestly have little idea in why this is the case that k = +- i does not work. It isn't like we are restricting k in R or something.

 

[superSAIyan2]

i meant if u sub x=i into the quartic u get -20i which is was Sy123 said

 

[Sy123]

Substitue 'i' into



???

I think he is referring to the polynomial in x (the actual question)

 

[RealiseNothing]

I think he is referring to the polynomial in x (the actual question)

Yer tbh I didn't read the question lol, it seemed to me he was referring to the equation I was using though so I went with it.

 

[superSAIyan2]

lol sorry if i was a bit hard to understand. but could any of you guys explain why you dont get 0 when u sub x = i into the quartic eqn

 

[RealiseNothing]

Also it's because the original polynomial has only 4 roots, and no more. Since we are letting k and -k be two roots, they can only take on one distinct value. If the 3, -3, i, -i were all possible values for k and -k, we would have a contradiction as there would be more than 4 roots. The roots would be 3, -3, i, -i, A, B, and hence there would be 6 roots, etc.

 

[superSAIyan2]

oh ok. so you are restricted with the number of solns you obtain as only some are correct - and to determine which roots are valid would you have to sub it into the eqn and check? Thanks realise and sy123