Math, asked by Asfiq6089, 1 year ago

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

Answered by mahendrachoudhary123
0

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

Answered by Anonymous
1

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

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