If the absolute value of the difference of roots of the equation $x^{2} + p x + 1 = 0$ exceeds $\sqrt{3 p}$

p < - 1 o r p > 4

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p > 4

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- 1 < p < 4

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0 \leq p < 4

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Solution

The correct option is D $p < - 1 o r p > 4$
Given equation , $x^{2} + p x + 1 = 0$
Let $α, β$ are the roots of the equation.
$\Rightarrow α + β = - p, α β = 1$
Given, $| α - β | > \sqrt{3 p}$
$\Rightarrow | α - β |^{2} > 3 p$
$\Rightarrow | α + β |^{2} - 4 α β > 3 p$
$\Rightarrow p^{2} - 3 p - 4 > 0$
$\Rightarrow (p - 4) (p + 1) > 0$
$\Rightarrow p \in (- \infty, - 1) \cup (4, \infty)$
$\Rightarrow p < - 1$ or $p > 4$ .

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If the absolute value of the difference of roots of the equation x2+px+1=0 exceeds √3p

The correct option is D p<−1orp>4Given equation , x2+px+1=0 Let α,β are the roots of the equation.⇒α+β=−p,αβ=1Given, |α−β|>√3p⇒|α−β|2>3p⇒|α+β|2−4αβ>3p⇒p2−3p−4>0⇒(p−4)(p+1)>0⇒p∈(−∞,−1)∪(4,∞) ⇒p<−1 or p>4.

If the absolute value of the difference of roots of the equation $x^{2} + p x + 1 = 0$ exceeds $\sqrt{3 p}$