Cubic equation (1 Viewer)

[member 6003]

i) x^3 + px + q = 0 has a root of multiplicity 2. Show 27q^2 + 4p^3=0
ii) Find the third root

I know how to do i) but I'm not sure what the answer is for ii) I managed to get +-2sqrt(p/3) as an answer but I'm pretty sure that's not what they're looking for.

 

[chilli 412]

Here's my working:
so if polynomial P(x) (lets just call it that) has two roots at x=2
we can write P(x) as (x-2)(x-2)(x-d) (where 'd' is the unknown third root) this will come in handy later
So for cubics in the form ax^3 + bx^2 +cx +d = 0,
And denoting the roots as α, β and γ,
we know that α + β + γ = -(b/a)
but P(x) is pretty much in the form of x^3 + 0x^2 + px + q
So our 'a' is equal to 1
our 'b' is equal to 0
And we know two of our roots α and β are both 2
so therefore 2 + 2 + γ = 0
4 + γ = 0
gamma, our third root, is equal to -4
so we've pretty much solved the second part of the question in this first part. which is strange but lets now write P(x)
P(x) = (x-2)(x-2)(x+4), now expanding we get:
P(x) = (x^2 -4x +4)(x+4)
P(x) = x^3 -12x +16
So therefore p = -12
And q = 16
Now substitute these values in for 27q^2 +4p^3
We get 27(16)^2 + 4(-12)^3 = 0
And we have pretty much shown that, sorry if there is an easier way for the first part
And now you know the second part as well, hope that helped

 

[member 6003]

Here's my working:
so if polynomial P(x) (lets just call it that) has two roots at x=2
we can write P(x) as (x-2)(x-2)(x-d) (where 'd' is the unknown third root) this will come in handy later
So for cubics in the form ax^3 + bx^2 +cx +d = 0,
And denoting the roots as α, β and γ,
we know that α + β + γ = -(b/a)
but P(x) is pretty much in the form of x^3 + 0x^2 + px + q
So our 'a' is equal to 1
our 'b' is equal to 0
And we know two of our roots α and β are both 2
so therefore 2 + 2 + γ = 0
4 + γ = 0
gamma, our third root, is equal to -4
so we've pretty much solved the second part of the question in this first part. which is strange but lets now write P(x)
P(x) = (x-2)(x-2)(x+4), now expanding we get:
P(x) = (x^2 -4x +4)(x+4)
P(x) = x^3 -12x +16
So therefore p = -12
And q = 16
Now substitute these values in for 27q^2 +4p^3
We get 27(16)^2 + 4(-12)^3 = 0
And we have pretty much shown that, sorry if there is an easier way for the first part
And now you know the second part as well, hope that helped

thanks

 

[Lith_30]

for part i

differentiate the polynomial

then we find the roots of the derivative

one of these roots will also have to be a root of the original function, so lets just sub it into the polynomial as it is.