Solve the equation.

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– Posted on 24/04/2014Posted in: MATH

Solve the equation
\left ( x+1 \right )^{4}+x^{4}=17

Write  x+1=t

t^{4}+\left ( t-1 \right )^{4}=17

t^{4}+\left ( t-1 \right )^{4}=16+1

t^{4}-1+\left ( t-1 \right )^{4}=16

\left ( t^{2}-1 \right )\left ( t^{2}+1 \right )+\left ( t-1 \right )^{4}=16

\left ( t-1 \right )\left ( t+1 \right )\left ( t^{2}+1 \right )+\left ( t-1 \right )\left ( t-1 \right )^{3}=16

\left ( t-1 \right )\left [ \left ( t+1 \right )\left ( t^{2}+1 \right )+\left ( t-1 \right )^{3} \right ]=16

\left ( t-1 \right )\left ( t^{3}+t^{2}+t+1+t^{3}-3t^{2}+3t-1 \right )=16

\left ( t-1 \right )\left ( 2t^{3}-2t^{2}+4t \right )=16

2t\left ( t-1 \right )\left ( t^{2}-t+2 \right )=16

t\left ( t-1 \right )\left ( t^{2}-t+2 \right )=8

\left ( t^{2}-t \right )\left ( t^{2}-t+2 \right )=8

Plug t^{2}-t=a

a\left ( a+2 \right )=8

a^{2}+2a-8=0

D=36, a_{1}=-4\wedge a_{2}=2

Go back to the replacement of  t

t^{2}-t=-4\Leftrightarrow t^{2}-t+4=0 no real solution D<0

OSE

t^{2}-t=2\Leftrightarrow t^{2}-t-2=0 has two solutions -1 and  2.

 

t=-1\Rightarrow x+1=-1\Rightarrow x=-2

t=2\Rightarrow x+1=2\Rightarrow x=1

The Solutions of the equations are A=\left \{ -2;1 \right \}.

 

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