Answer to Question #152751 in Algebra for Hardy

Question

Solve (x^2-11x+29)^(6x^2+x-2)=1

Answer 1

$$x^2-11x+29=1$$

$$x^2-11x+29-1=0$$

$$x^2-11x+28=0$$

$$(x-4)(x-7)=0$$

$$x_1=4, x_2=7$$

Or

$$6x^2+x-2=0, x^2-11x+29\not=0$$

$$(3x+2)(2x-1)=0$$

$$x_3=-\dfrac{2}{3}, x_4=\dfrac{1}{2}$$

Let

$$x^2-11x+29=-1$$

$$x^2-11x+29+1=0$$

$$x^2-11x+30=0$$

$$(x-5)(x-6)=0$$

$$x=5\ or\ x=6$$

If $x=5$

$$6x^2+x-2=6(5)^2+5-2=153$$

$$(-1)^{153}\not=1$$

If $x=6$

$$6x^2+x-2=6(6)^2+6-2=220$$

$$(-1)^{220}=1$$

$$x_5=6$$

$\bigg\{-\dfrac{2}{3}, \dfrac{1}{2}, 4, 6,7\bigg\}$