Math, asked by guksimgn, 4 days ago

find the condition that the equation x^3+3px^2+3qx+r=0 maybe in G.P​

Answers

Answered by poorvashmk0611
1

Answer:

Given that the roots of x

3

+3px

2

+3qx+r=0 are in H.P. (1)

Let x=

x

1

⇒(

x

1

)

3

+3p(

x

1

)

2

+3q(

x

1

)+r=0

Multiplying by x

3

throughout; we get

x

3

(

x

1

)

3

+x

3

3p(

x

1

)

2

+x

3

3q(

x

1

)+rx

3

=0

⇒rx

3

+3qx

2

+3px+1=0 (2)

The roots of the equation (2) being reciprocal of the roots of the equation (1) must be in A.P.

Let the roots of eq. (2) be α−β,α,α+β

∴ from equation (2); we get

sum of the roots=α−β+α+α+β=−

r

3q

⇒3α=−

r

3q

⇒α=−

r

q

Since α is a root of (2)

∴r(

r

−q

)

3

+3q(

r

−q

)

2

+3p(

r

−q

)+1=0

r

2

−q

3

+

r

2

3q

3

r

3pq

+1=0

⇒2q

3

−3pqr+r

2

=0

∴ The required condition is 2q

3

−3pqr+r

2

=0

Answered by lathagowda301
0

Answer:

Given that the roots of x

3

+3px

2

+3qx+r=0 are in H.P. (1)

Let x=

x

1

⇒(

x

1

)

3

+3p(

x

1

)

2

+3q(

x

1

)+r=0

Multiplying by x

3

throughout; we get

x

3

(

x

1

)

3

+x

3

3p(

x

1

)

2

+x

3

3q(

x

1

)+rx

3

=0

⇒rx

3

+3qx

2

+3px+1=0 (2)

The roots of the equation (2) being reciprocal of the roots of the equation (1) must be in A.P.

Let the roots of eq. (2) be α−β,α,α+β

∴ from equation (2); we get

sum of the roots=α−β+α+α+β=−

r

3q

⇒3α=−

r

3q

⇒α=−

r

q

Since α is a root of (2)

∴r(

r

−q

)

3

+3q(

r

−q

)

2

+3p(

r

−q

)+1=0

r

2

−q

3

+

r

2

3q

3

r

3pq

+1=0

⇒2q

3

−3pqr+r

2

=0

∴ The required condition is 2q

3

−3pqr+r

2

=0

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