Polynomial Question - HSC 2000 (1 Viewer)

[rsingh]

Hey guys,

I was doing some random questions from past papers, and I came across this question which I'm having trouble to do. It's from the HSC in 2000.

The polynomial P(x) = x^3 + px^2 + qx + r has real roots, ROOT k, - ROOT K and α .

(i) Explain why α + p = 0.
(ii) Show that kα = r .
(iii) Show that pq = r .​

If anyone could help, it would be greatly apprecaited. Thanks.

 

[prsce24]

i) from relation between roots and coeff
root k + - root k + a = -coeffx^2/coeffx^3 = -p
a=-p
a+p=0

ii) from relation....
(root k x - root k) + ( root k x a) + ( - root x a) = coeffx^1/coeffx^3 gives -k=q

taking three roots at a time : (rootk x -rootk x a) = -coeffx^0/coeffx^3 = -r
gives -ka=-r
hence ka=r

iii)
from i) and ii)

ka=r
(-q)(-p)=r since (-k=q and a+p=0)
hence pq=r

 

[currysauce]

ok, remember your root rules.

that polymonial is a cubic polymonial, indicated by the degree of 3.

now P(x) = x^3 + px^2 + qx + r


Let roots be root(k), -root(k), and a (APLHA).

(i) a + p = 0 sum of the roots, -b/a which is -p/1

so we have
root (k) + - root(k) + a = -p/1 <--- now the root(k)'s cancel and -p/1 = -p

so we have a = -p which then simplifies to a + p = 0

(ii) ka = r now this is product of the roots, which is -d/a

so we have
root(k) x -root(k) x a = -r/1 then gives:

-ka = -r (divide both sides by -1)
ka = r

(iii) ok lets look at the above equation -ka = -r

we know from (i) that a = -p
and from sum of the roots (TWO at a time, that is d/a, we have):

a root(k) - a root(k) -k = q/1

so we have -k = q which is the same as k = -q

so now we combine k and a -( a x k ) = -r

- (pq) = -r
so pq=r




HOPE THIS HELPS!!

 

[rsingh]

Thanks guys.
Highly appreciated!