Find the condition that the equation x^3-px^2+qx-r=0 should have its roots in GP
Answers
Answered by
0
Answer:
Given, x
3
−px
2
+qx−r=0
Case 1: Roots are α,−α,β
Sum of the roots, S
1
=α−α+β=p⟹β=p
Sum of the roots taken 2 at a time, S
2
=−α
2
+αβ−αβ=q⟹−α
2
=q
Product of the roots, $$S_3 = -\alpha^2\beta = r
From the above equations we can see that S
1
⋅S
3
=S
2
⟹pq=r
Case 2: Roots are in geometric progression. Let the roots be α,β,γ
∴
β
α
=
γ
β
⟹β
2
=αγ⟹β
3
=αβγ=r
Since, β is the roots of the given equation, we have
β
3
−pβ
2
+qβ−r=0⟹r–p⋅r
2/3
+q⋅r
1/3
–r=0
⟹p
3
r=q
3
[taking cubes on both sides and simplifying]
Similar questions