Let us asume

and

are distinct roots of x
2 + px + q = 0.

and

are real and distinct roots, discriminant D > 0.

p
2 – 4q > 0 …..(3)
4 and
4 are roots of x
2 – rx + s = 0.
4 +
4 = (
2 +
2)
2 – 2
2
2 = [(
+ 
)
2 – 2


]
2 – 2(


)
2 = (p2 – 2q)2 – 2q2 = p4 – 4p2q + 4q2 – 2q2
= p4 – 4p2q + 2q2 = p2(p2 – 4q) + 2q2 [From (1) & (2)]

r = p
2(p
2 – 4q) + 2q
2 & s = q
2
4 +
4 > 0 ;
4
4 > 0 for all
,
i.e. r = p
2(p
2 – 4q) + 2q
2 > 0 ; s>0
Now, consider x4 – 4qx + 2q2 – r = 0
D = (-4q)
2 – 4(2q
2 – r) = 4(4q
2 – 2q
2 + r) = 4(2q
2 + r) > 0

q
2 > 0 & r >0
This shows that roots of x4 – 4qx + 2q2 – r = 0 are real.
Let f(x) = x4 – 4qx + 2q2 – r
f(0) = 2q2 – r = 2q2 – [p2(p2 – 4q) + 2q2] = 2q2 – p2(p2 – 4q) – 2q2
= – p2(p2 – 4q)
From (3), p
2 – 4q > 0

p
2(p
2 – 4q) > 0

– p
2(p
2 – 4q)
i.e. f(0)
Graph of f(x) is concave upwards (because coefficient of x2 is 1 > 0)
with the value of f(x) at x = 0 negative.
This is possible when one root is positive while other is negative.