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
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!