# Finding the roots of cubic equations

• Mar 20th 2009, 08:18 AM
s_cSM
Finding the roots of cubic equations
Hi, I've tried the question below and got stuck halfway =) I'd really appreciate any help!! Thanks!!

QUESTION: ax^3+bx^2+cx+d=0 has roots A, B and C. Find the new equation for roots (A-k), (B-k), (C-k).

From eq given,
SOR=A+B+C=-b/a
Sum of product of roots taking 2 at a time=AB+BC+AC=c/a
POR=ABC=-d/a

Using roots (A-k), (B-k), (C-k),

SOR=A+B+C-3k=-b/a -3k

Sum of product of roots taking 2 at a time
=(A-k)(B-k)+(B-k)(C-k)+(A-k)(C-k)
=AB+BC+AC+(-A-B-B-C-A-C)k+3k^2
=AB+BC+AC+(-2A-2B-2C)k+3k^2
=c/a -2k(A+B+C)+3k^2
=c/a -2k(-b/a)+3k^2
=c/a +(2bk)/a + 3k^2

POR
=(A-k)(B-k)(C-k)
=ABC+(-AB-AC-BC)k+(A+B+C)k^2-k^3
=-d/a -k(c/a) + (b/a)k^2 - k^3

After this, I got stuck. How do I get the cubic equation? Something just seems wrong. Help! =) Thanks everyone!
• Mar 20th 2009, 08:34 AM
running-gag
Hi

The equation is therefore

x^3+(b/a + 3k)x^2+(c/a +(2bk)/a + 3k^2)x+(d/a + k(c/a) + (b/a)k^2 + k^3)=0

I think that you have made a mistake in the last part
POR
=(A-k)(B-k)(C-k)
=ABC+(-AB-AC-BC)k+(A+B+C)k^2-k^3
=-d/a -k(c/a) - (b/a)k^2 - k^3
• Mar 20th 2009, 08:54 AM
s_cSM
Thanks running-gag for pointing out that mistake! (Nod)

The answer given in my book is
a(x+k)^3 +b(x+k)^2 +c(x+k) +d=0

how did they exactly get that answer?

are we supposed to factorize or do something along that line?
• Mar 20th 2009, 09:06 AM
running-gag
Suppose that A is a solution of ax^3+bx^2+cx+d=0
then aA^3+bA^2+cA+d=0

a[(A-k)+k]^3+b[(A-k)+k]^2+c[(A-k)+k]+d=0

Therefore A-k is solution of a[x+k]^3+b[x+k]^2+c[x+k]+d=0
• Apr 18th 2009, 07:16 PM
noobonastick
Quote:

Originally Posted by running-gag
Suppose that A is a solution of ax^3+bx^2+cx+d=0
then aA^3+bA^2+cA+d=0

a[(A-k)+k]^3+b[(A-k)+k]^2+c[(A-k)+k]+d=0

Therefore A-k is solution of a[x+k]^3+b[x+k]^2+c[x+k]+d=0

So basically all that working out was redundant?