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Grade:11

1 Answers

Samyak Jain
333 Points
5 years ago
Let us asume \alpha and \beta are distinct roots of x2 + px + q = 0.
\therefore \alpha + \beta = – p  ...(1)  &  \alpha\beta = q   ...(2)
\because \alpha and \beta are real and distinct roots, discriminant D > 0.
\Rightarrow p2 – 4q > 0                   …..(3)
\alpha4 and \beta4 are roots of x2 – rx + s = 0.
\therefore \alpha4 + \beta4 = r  &  \alpha4\beta4 = s
\alpha4 + \beta4 = (\alpha2 + \beta2)2 – 2\alpha2 \beta2 = [(\alpha + \beta)2 – 2\alpha\beta]2 – 2(\alpha\beta)2 
              = (p2 – 2q)2 – 2q2 = p4 – 4p2q + 4q2 – 2q2 
              = p4 – 4p2q + 2q2 = p2(p2 – 4q) + 2q2                [From (1) & (2)]
\alpha4\beta4 = (\alpha\beta)2 = q2
\therefore r = p2(p2 – 4q) + 2q2    &    s = q2
\alpha4 + \beta4 > 0 ; \alpha4\beta4 > 0 for all \alpha,\beta  i.e.  r = p2(p2 – 4q) + 2q2 > 0  ;  s>0
Now, consider x4 – 4qx + 2q2 – r = 0
D = (-4q)2 – 4(2q2 – r) = 4(4q2 – 2q2 + r) = 4(2q2 + r) > 0  \because q2 > 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), p2 – 4q > 0  \Rightarrow p2(p2 – 4q) > 0  \Rightarrow – p2(p2 – 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.

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