if the roots of the equation x^3+3px^2+3px+r=0 be in H.P then show that 2q^3=r(3pq-r).
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2
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ANSWER
Given equation x
3
+3px
2
+3qx+r=0
Let roots of the equation be α,β,γ
Since the roots are in harmonic progression,
α
1
+
γ
1
=
β
2
⇒αβ+βγ=2αγ ....(1)
⇒αβ+βγ+αγ=3q
⇒3αγ=3q[From (1)]
⇒αγ=q .....(2)
αβγ=−r⟹qβ=−r⟹β=−
q
r
[from (2)]
Since β is a root of the given equation, therefore
β
3
+3pβ
2
+3qβ+r=0
⇒(−
q
r
)
3
+3p(−
q
r
)
2
+3q(−
q
r
)+r=0
⇒−r
2
+3pqr−3q
3
+q
3
→2q
3
=3r(pq−r)
Answered by
1
Answer:
sorry I can't understand this question
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