How can I solve this equation for real numbers? How can I solve this equation for real numbers?
$$(x+2)^4+x^4=82.$$
I tried $(x+2)^4+x^4-82=
2x^4 + 8 x^3 + 24 x^2 + 32 x - 66=0$
It is very difficult to continue.
 A: $$u=x+1$$
$$(u+1)^4+(u-1)^4=82$$
$$u^4+4u^3+6u^2+4u+1+u^4-4u^3+6u^2-4u+1=82$$
$$2u^4+12u^2=80$$
$$u^4+6u^2-40=0$$
$$(u^2+10)(u^2-4)=0$$
$$u^2=4$$
$$u=\pm2$$
$$x=-3\lor x=1$$
A: Look my method:
Let $x+1=y$
$(y+1)^4+(y-1)^4=82 \Rightarrow y^4+6y^2-40=0 \Rightarrow$ $y^2=-10$ and $y^2=4$
Finally, $x_1=-3, x_2=-2$
A: Set $y = x + 1$. The equation rewrites as $(y + 1)^2 + (y - 1)^2 = 82$, that is $$2(y^4 + 6y^2 + 1) = 82 \iff y^4 + 6y^2 - 40 = 0$$ Now let $z = y^2$: the equation has become $z^2 + 6z - 40 = 0$, which is a simple quadratic equation with solutions $$z_{1, \: 2} = -3 \pm \sqrt{9 + 40} = -3 \pm 7 \implies z_1 = 4, \; z_2 = -10$$ Since $z = y^2 \ge 0$, $z_2$ is not acceptable, so $z = 4$. Finally, substituting backwards we get $$x = y - 1 = \pm\sqrt{z} - 1 = \pm 2 - 1$$ Thus there are two real solutions, namely $x_1  = 1$ and $x_2 = -3$.
A: Let $x+2=a$ and $x=b$.
Thus, $a-b=2$ and we have
$$a^4+b^4=82$$ or
$$(a^2+b^2)^2-2a^2b^2=82$$ or
$$((a-b)^2+2ab)^2-2a^2b^2=82$$ or
$$(4+2ab)^2-2a^2b^2=82$$ or $$a^2b^2+8ab-33=0,$$ which gives
$ab=-11$, $x(x+2)+11=0,$ which is impossible or
$ab=-3$, $x(x+2)=3,$ which gives the answer:
$$\{1,-3\}.$$
We can end also your way.
We need to solve
$$2x^4 + 8 x^3 + 24 x^2 + 32 x - 66=0$$ or
$$x^4+4x^3+12x^2+16x-33=0$$ or
$$x^4-x^3+5x^3-5x^2+17x^2-17x+33x-33=0$$ or
$$(x-1)(x^3+5x^2+17x+33)=0$$ or
$$(x-1)(x^3+3x^2+2x^2+6x+11x+33)=0$$ or
$$(x-1)(x+3)(x^2+2x+11)=0,$$
which gives the same answer.
A: A different viewpoint, in case you don't like the (smart) change of variables $y:=x+1$:
First a bit of calculus: Observe that $f(x):=(x+2)^4+x^4-82$ is a twice-differentiable function, with second derivative $f''(x)=12(x+2)^2+12x^2$, which is positive everywhere. Hence $f$ is convex, so it has at most two real roots. Therefore there are at most two different values such that $(x+2)^4+x^4=82$. By Descartes' sign of rules we deduce that there is exactly one positive real solution, while by the same rule applied to $f(-x)$ we find that there are either 3 or 1 negative real solutions; therefore we have two solutions, one positive and one negative.
Now a bit of playing-with-numbers: Observe that $82=81+1=3^4+1^4$. Therefore
$$(x+2)^4+x^4 = 3^4+1^4 = (1+2)^4 + 1^4$$
gives us the positive solution $x=1$. For the negative solution, due to the fact that $x^4=(-x)^4$, we can try with $x=-1$ and $x=-3$: the first one is a no-go, but the second one satisfies $x+2=-1$, so that
$$(x+2)^4+x^4 = (-1)^4 + (-3)^4 = 1^4 + 3^4 = 82.$$
Hence the solutions are $x=1$ and $x=-3$.
A: Two solutions of $82=x^4+(x+2)^4$ are given by $x=1$ and $x=-3$. It follows that
$$ x^4+(x+2)^4-82 = (x-1)(x+3) q(x) $$
where $q(x)$ is a quadratic polynomial. Find it, solve $q(x)=0$ and you have the complete set of solutions ($q(x)=2\left[(x+1)^2+10\right]$, so the only real solutions are the trivial ones).
