Quadratic Equation and Inequalities

Since α, β are the roots of equation, we have
x2+2px+q = 0
α + β = - 2p , αβ = q
and α ,

1

β

are the roots of equation
ax2+2bx+c = 0
α+

1

β

=

−2b

a

,

α

β

=

c

a

If β = 1 ; α = q =

x

a

⇒ c = qa (not possible)
and α + 1 = - 2 p =

−2b

a

⇒ b = pq (not possible)
Therefore Statement 2 is correct.
Now, let as consider the roots to be imaginary:
β =

−

α

,

1

β

=

−

α

⇒ β =

1

β

(not possible)
Hence, the roots are real in both equations. Hence,
(4p2−4q)(4b2−4ac) ≥ 0
(p2−q)(b2−ac) ≥ 0