Question
If $a,b$ are the roots of the quadratic equation $x^{2} + \lambda x - \frac{1}{2\lambda^{2}} = 0$, where $\lambda$ is a real parameter. Then the minimum value of $a^{4} + b^{4}$ is
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Solution
$a^{4} + b^{4} = \left( a^{2} + b^{2} \right)^{2} - 2a^{2}b^{2} = \left( (a + b)^{2} - 2ab \right)^{2} - 2a^{2}b^{2}$
$$\begin{matrix} & \ = \left( \lambda^{2} + \frac{1}{\lambda^{2}} \right) - \frac{1}{2\lambda^{4}} \\ & \ = \lambda^{4} + \frac{1}{2\lambda^{4}} + 2 \\ & \ \geq 2 + \sqrt{2} \end{matrix}$$
19, If $a,b,c$ are complex numbers and $a + b + c = ab + bc + ca = abc = 1$, then find $a + b^{4} + c^{4}$
(A) 1
(B) 0
(C) 3
(D) 8
Sol. $\begin{matrix} & x^{3} - x^{2} + x - 1 = 0\sum_{c}^{a}\mspace{2mu}\mspace{2mu} b \\ & \ \Rightarrow \ \left( x^{2} + 1 \right)(x - 1) = 0 \\ & \ \Rightarrow \ (x = 1,i, - i \end{matrix}$
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