Let a,b be the roots of x²-x+p=0 and c,d be the roots of x²-4x+q=0 . If a,b,c,d are in GP, then integral values of p and q will be ???????
Answers
Answer:
b/a=c/b=d/c=k
a+b=1. a=1/(1+k)
ab=p a*a*k=p. k/(k+1)'2=p. 1/k +k +2 =1/p then
c+d=4. c=4/(1+k)
cd=q. c*c*k=q. 16/q =1/k +k +2
16/q =1/p
then q=-16 and p =-1
note 2<=k+1/k<=-2
Answer:
p=-2 and q=-32
Step-by-step explanation:
a,b, c,d are in GP. assume that common ratio=r;
then b=a.r; c=a.r^2; d=a.r^3;
From the equation x^2-x+p=0, ----(1)
sum of the roots a+b=1
a(1+r)=1....................((2)
From the equation x^2-4x+q=0--------------(3)
sum of the roots c+d=4
a.r^2.(1+r)=4--------------------(4)
Divide (4) by (2) we get
r^2=4
r=2 or -2
from(1) we get
a=1/3 or -1
for p, q to be integers a= -1
So from (2)
(-1)(1+r)=1
1+r=-1
r=-2
so a=-1,
b=-1*(-2)=2
c=-1(-2)^2=-4
d=--(-2)^3=-1*-8=8
Now product of roots of x^2-x+p=0
ab=(-1)(2)=p
p=-2
and Now product of roots of x^2-4x+q=0
cd=(-4)(8)=-32=q
so q=-32
hence p=-2 and q=-32