Question

If no roots of the equation x2 - px + 1 = 0 is real, then​

Answer 1

answer : the value of p ∈ (-2, 2)

we have to find value of p for which no roots of equation x² - px + 1 is real.

for no real roots , D = b² - 4ac < 0

⇒(-p)² - 4(1)(1) < 0

⇒p² - 4 < 0

⇒(p - 2)(p + 2) < 0

⇒-2 < p < 2

hence value of p ∈ (-2, 2) for which no roots of given equation is real.

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Answer 2

Answer:

| P | > 2

Step-by-step explanation:

d= b²- 4ac>0

= (-p)² - 4(1)(1) >0

= p² - 4>0

=p²>4

squre root both sides

=p>√4

=p>±2

=| p |>2