lf $p, q, r \in R$ and the quadratic equation $p x^{2} + q x + r = 0$ has no real root, then

p (p + q + r) > 0

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p + q + r = 0

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q (p + q + r) > 0

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p + q + r > 0

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Solution

The correct option is A $p (p + q + r) > 0$
$∵ p x^{2} + q x + r h a s n o r e a l r o o t s t h e n t h e r o o t s a r e ω, ω^{2} . W h e n ω^{3} = 1 a n d 1 + ω + ω^{2} = 0. ∴ ω + ω^{2} = - \frac{q}{p} \Rightarrow - 1 = - \frac{q}{p} \Rightarrow p = q . . . . . (1) A g a i n, ω . ω^{2} = \frac{r}{p} \Rightarrow 1 = \frac{r}{p} \Rightarrow r = p . . . . . . (2) ∴ F r o m (1) a n d (2) p = q = r . . . . . . (3) N o w, p > 0 ∴ p (p + q + r) = p (3 p) [f r o m (3)] = 3 p^{2} > 0 (∵ p > 0) ∴ p (p + q + r) > 0 A n s O p t i o n A .$

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If $p x^{2}$ + qx + r = 0 has no real roots and p, q, r are real such that p + r > 0 then

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a

p x^{2} + q x + r = 0

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