Question 31432: Assume that a and b are different zeros of p(x)= x^3+kx-1. Show that ab is a zero of q(x)=x^3-kx-1 where k is any real number. Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! X^3+KX-1=0.....A AND B ARE 2 ROOTS.SINCE THIS IS A CUBIC IN X ,THERE
WILL BE 3 ROTS .LET THE THIRD ROOT BE C.SO WE KNOW THAT...
FOR X^3+0*X^2+KX-1=0
S1=A+B+C=0....OR...C= -(A+B)....................I
S2=AB+BC+CA=K.....OR.....AB+C(A+B)=K..............II
S3=ABC=1.....OR......C=1/AB...........................................................III
SUBSTITUTING -C FOR (A+B) IN EQN.II FROM EQN.I WE GET
AB-C*C=K.....SUBSTITUTING FOR C FROM EQN.III WE GET....
AB-1/(AB)^2-K=0.............MULTIPLYING BY (AB)^2..WE GET
(AB)^3-1-K(AB)^2=0........................................................IV
PUTTING X IN PLACE OF AB WE GET
X^3-KX^2-1=0.......IS THE EQN. SATISFIED BY AB IN VIEW OF WHAT WE
PROVED IN EQN.IV
