If all the roots of the equation x3+px+q=0,p,q∈R,q≠0 are real, then which the following is correct?
A
p<0
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B
p=0
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C
p>0
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D
p≤0
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Solution
The correct option is Ap<0 Let α,β,γ be the roots of the given equation. Then, α+β+γ=0αβ+βγ+γα=pαβγ=−q
Now, p=αβ+βγ+γα ⇒p=12[(α+β+γ)2−(α2+β2+γ2)] ⇒p=−12(α2+β2+γ2)
As q≠0, so αβγ≠0⇒α≠0,β≠0,γ≠0
Therefore, α2+β2+γ2>0⇒−12(α2+β2+γ2)<0⇒p<0