Answer

Find the midpoint between 2 and 3

m = \dfrac{2+3}{2} =\dfrac{5}{2}

Then

\Big( x-\dfrac{5}{2} +\dfrac{1}{2} \Big) ^4+\Big( x-\dfrac{5}{2} -\dfrac{1}{2} \Big) ^4=1

Let

u = x - \dfrac{5}{2}
(1)

Then

The equation is transformed to

\Big( u +\dfrac{1}{2} \Big) ^4+\Big( u-\dfrac{1}{2} \Big) ^4=1

\Big( u^2+u+\dfrac{1}{4} \Big) ^2+\Big( u^2-u+\dfrac{1}{4} \Big) ^2 =1

2\Big( u^2+ \dfrac{1}{4}\Big) ^2+2u^2 = 1

2(u^4+\dfrac{u^2}{2}+\dfrac{1}{16} )+2u^2=1

2u^4+3u^2-\dfrac{7}{8} = 0

16u^4+24u^2-7=0

(4u^2+7)(4u^2-1) = 0

\begin{cases} u_1 &= \dfrac{1}{2} \\ u_2 &= -\dfrac{1}{2} \end{cases}

or

\begin{cases} u_3 &= \dfrac{\sqrt{7} }{2}i \\ u_4 &= -\dfrac{\sqrt{7} }{2}i \end{cases}

Substitute u back to (1)

x = u+\dfrac{5}{2}

Then

x_1 = \dfrac{1}{2} +\dfrac{5}{2} = 3

x_2 = -\dfrac{1}{2} +\dfrac{5}{2} = 2

x_3 = \dfrac{5}{2}+\dfrac{\sqrt{7} }{2}i

x_4 = \dfrac{5}{2} - \dfrac{\sqrt{7} }{2}i


Steven Zheng posted 18 hours ago

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