Solve the equation.

Categories MATH
Date 07/08/2022
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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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