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- Mimi_x3

The polynomial P(x) = x^3 - 2x^2 + kx + 24 has roots a , b and γ
(i) It is known that two of the roots are equal in magnitude but opposite in sign.
Find the third root and hence find the value of k
Can you please help .

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- Mimi_x3

- katieb

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- anonymous

:@

- Mimi_x3

:@ ?? why yu angry ? LOL

- anonymous

3rd root is 2

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- anonymous

not angry

- anonymous

just having a bad day

- Mimi_x3

how is 3rd root 2 ?

- Mimi_x3

lols , having a bad day ? why ?

- anonymous

I passed the entry test to uni, but apparently, I need to get minimum of 2 A levels to get into it. A levels take 2 years to get. I have 4 months.

- Mimi_x3

ohh ok LOL

- anonymous

I got so far (for roots a,-a and c)
3a +c = 2 and
a^2 c = 24
U get that?

- Mimi_x3

yeah

- Mimi_x3

w8 , what's c ?

- anonymous

I don't mean understand, I mean u calculated same?

- anonymous

I said for roots, a,-a and c

- Mimi_x3

ohh lols , i dnt even know how to do it

- anonymous

There is similar (not same) formula like Vieta for cubic as for quadratic....

- Mimi_x3

What's Vieta ?

- anonymous

Formulae for sum and product of roots of quadratic.

- Mimi_x3

ohh that

- anonymous

Let a be the first root >0. Then b=-a
Now since a cubic function with 3 real roots then must be of the form:
(x-a)(x-b)(x-v)= x^3 - 2x^2 + kx + 24
But b=-a so LHS is (x-a)(x+a)(x-v)=(x^2-a^2)(x-v)
Multiplying the left hand side out we get:
=x^3-v*x^2-a^2*x+v*a^2=P(x)
From this we see that v=2 which implies 24=2*a^2
then a^2=12 but -a^2*x=kx
thus k=-12

- Mimi_x3

thank you (:

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