Question

73. If the absolute value of the difference of roots of the equation
x? + px + 1 = 0 exceeds v3p then

Answer 1

Answer:

Given equation , x²+px+1=0

Let α,β are the roots of the equation.

⇒α + β = − p, αβ = 1

Given, ∣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

Hope it will be helpful :)

Answer 2

Answer:

refers to the attachment