Polynomials Question. (1 Viewer)

[Divinity_]

The polynomial P(x) = x^3 + px^2 + qx + r = 0 has roots sqrt(k), -sqrt(k), and a.

1. Explain why a + p = 0
2. Show that ka = r
3. Show that pq = r

Source:
Question 4 (c)

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/00mathematics34U.pdf

 

[study-freak]

P(x) = x^3 + px^2 + qx + r = 0
1.Sum of roots=sqrt(k)-sqrt(k)+a=-p
a+p=0

2.Product of roots=-ka=-r
ka=r

3.Sum of roots taken 2 at a time=-k-a sqrt(k)+a sqrt(k)=q
k=-q
using 2and 1, ka=r=-qa=-q(-p)=pq
pq=r

 

[lyounamu]

The polynomial P(x) = x^3 + px^2 + qx + r = 0 has roots sqrt(k), -sqrt(k), and a.

1. Explain why a + p = 0
2. Show that ka = r
3. Show that pq = r

Source:
Question 4 (c)

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/00mathematics34U.pdf

1. addition of all roots = -b/a
a = -p
a + p = 0 ...(1)

2. multiples of all roots = -d/a
-ka = -r
ka = r ...(2)

3. addition of multiples of two roots at a time = c/a
-k = q
- r/a = q from 2
r/p = q from 1
pq = r

 

[lychnobity]

The polynomial P(x) = x^3 + px^2 + qx + r = 0 has roots sqrt(k), -sqrt(k), and a.

1. Explain why a + p = 0
2. Show that ka = r
3. Show that pq = r

Source:
Question 4 (c)

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/00mathematics34U.pdf

1) sum roots = -p
ie √k + (-√k) + a = -p
a = -p
.'. a + p = 0 ... 1

2) product of roots = -r
√k .(-√k).a = -r

-ka = -r, ka = r ... 2

3) sum of roots two at a time = q

a√k + a(-√k) + √k(-√k) = q

-k = q... 3

From eqn 1, a = -p let this eqn be 4

eqn 3 x eqn 4

ak = pq, but ka = r (from eqn 2)

therefore, r = pq

 

[Revacious]

i did b and c, and was stuck on a for about 4 minutes.

my working was

-p = a

and i was like how the hell do you do it?

self esteem destroyer

 

[lychnobity]

i did b and c, and was stuck on a for about 4 minutes.

my working was

-p = a

and i was like how the hell do you do it?

self esteem destroyer

lol add p to both sides