Quadratic EquationHard
Question
The value of maximum real root minus the minimum real root of the equation $\left( x^{2} - 5 \right)^{4} + \left( x^{2} - 7 \right)^{4} = 16$ is
Options
A.$\sqrt{5} + \sqrt{7}$
B.$2\sqrt{5}$
C.$\sqrt{28}$
D.$4\sqrt{2}$
Solution
Put $x^{2} - 6 = t$
$$\begin{matrix} & & (t + 1)^{4} + (t - 1)^{4} & = 16 \\ \Rightarrow & t^{4} + 6t^{2} - 7 & & = 0 = \left( t^{2} + 7 \right)\left( t^{2} - 1 \right) \\ \Rightarrow & x^{2} - 6 & & = \pm 1 \\ \Rightarrow & x & & = \pm \sqrt{7}, \pm \sqrt{5} \end{matrix}$$
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