Solve the equation

$$
\sqrt[4]{2-x}+\sqrt[4]{15+x}=3
$$
I. solution. Squaring both sides and then rearranging:

$$
\sqrt{2-x}+\sqrt{15+x}=9-2 \sqrt[4]{(2-x)(15+x)}
$$

Squaring again:

$$
\begin{gathered}
(2-x)+(15+x)+2 \sqrt{(2-x)(15+x)}= \\
=81+4 \sqrt{(2-x)(15+x)}-36 \sqrt[4]{(2-x)(15+x)}
\end{gathered}
$$

Let $a=\sqrt[4]{(2-x)(15+x)}$. Then

$$
\begin{aligned}
17+2 a^{2} & =81+4 a^{2}-36 a \\
0 & =2 a^{2}-36 a+64
\end{aligned}
$$

From this, $a=16$ or $a=2$.

If $a=16$, then the equation $16=\sqrt[4]{(2-x)(15+x)}$ raised to the fourth power on both sides and rearranged gives the equation $0=x^{2}+13 x+65506$. This has a negative discriminant, so it has no solution.

If $a=2$, then the equation $2=\sqrt[4]{(2-x)(15+x)}$ raised to the fourth power on both sides and rearranged gives the equation $0=x^{2}+13 x-14$. The solutions to this are $x_{1}=-14$ and $x_{2}=1$, both of which are solutions to the original equation as well.

II. solution. Let $a=\sqrt[4]{2-x}, b=\sqrt[4]{15+x}$. Then $a+b=3$, and $a^{4}+b^{4}=17$. Write out the fourth power of $(a+b)$ and transform it:

$$
(a+b)^{4}=a^{4}+4 a^{3} b+6 a^{2} b^{2}+4 a b^{3}+b^{4}=17+4 a^{3} b+6 a^{2} b^{2}+4 a b^{3}
$$

Using the fact that $(a+b)^{4}=81$, we get the following equation:

$$
32=2 a^{3} b+3 a^{2} b^{2}+2 a b^{3}=a b\left(2 a^{2}+3 a b+2 b^{2}\right)=a b\left(2(a+b)^{2}-a b\right)=a b(18-a b)
$$

Expanding the parentheses and reducing to zero:

$$
(a b)^{2}-18 a b+32=0
$$

This is a quadratic equation in $a b$, with solutions: $a b=16$, and $a b=2$. From these, using $a=3-b$: $(3-b) b=16$, or $(3-b) b=2$. Thus, we get two quadratic equations in $b$. The first has a negative discriminant, so it has no real roots. From the second equation: $b_{1}=2, b_{2}=1$. Hence: $\sqrt[4]{15+x}=2, x_{1}=1 ; \sqrt[4]{15+x}=1, x_{2}=-14$. Both values of $x$ are solutions to the original equation as well.

III. solution. Let

$$
f(x)=\sqrt[4]{2-x}+\sqrt[4]{15+x}
$$

where $x \in[-15,2]$. Perform a function analysis. First, calculate the first derivative:

$$
\begin{aligned}
f^{\prime}(x) & =\frac{1}{4} \cdot \frac{1}{\sqrt[4]{(2-x)^{3}}} \cdot(-1)+\frac{1}{4} \cdot \frac{1}{\sqrt[4]{(15+x)^{3}}} \cdot(+1)= \\
& =\frac{1}{4} \cdot\left(\frac{1}{\sqrt[4]{(15+x)^{3}}}-\frac{1}{\sqrt[4]{(2-x)^{3}}}\right)
\end{aligned}
$$

This can only be 0 if $x \neq-15$ and $x \neq 2$ and

$$
\begin{gathered}
\frac{1}{\sqrt[4]{(15+x)^{3}}}=\frac{1}{\sqrt[4]{(2-x)^{3}}} \\
15+x=2-x, \quad x=-6.5
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_67bca1062d9e2c6bbf7cg-1.jpg?height=363&width=404&top_left_y=2345&top_left_x=855)

Using the fact that $f^{\prime}(x)$ is defined throughout the interval $]-15 ; 2[$, and for example, $f^{\prime}(-7)>0$, and $f^{\prime}(0)>3$. Therefore, the function $f$ takes the value 3 at two points in the given interval. Since $f(1)=f(-14)=3$, and $f(-15)=f(2) \approx 2.03$, the equation has two solutions: $x=1$ and $x=-14$.

Remark. Plotting the two terms of $f(x)$ separately, it can be inferred from the graph that the sum of the two functions will be maximal at the intersection point $(x=-6.5)$, since the two functions are symmetric about the intersection point, and for example, to the right of the intersection point, $\sqrt[4]{2-x}$ decreases "faster" than $\sqrt[4]{15+x}$ increases.

![](https://cdn.mathpix.com/cropped/2024_05_02_67bca1062d9e2c6bbf7cg-2.jpg?height=398&width=423&top_left_y=533&top_left_x=840)

Indeed, since the function $g(x)=\sqrt[4]{x}$ is strictly concave, the value of the function $f$ at $-6.5$ is smaller, or larger, for $x$ values:

$$
\begin{aligned}
f(-6.5 & +\varepsilon)=f(-6.5-\varepsilon)=\sqrt[4]{8.5-\varepsilon}+\sqrt[4]{8.5+\varepsilon}= \\
& =2 \cdot[0.5 \cdot \sqrt[4]{8.5-\varepsilon}+0.5 \cdot \sqrt[4]{8.5+\varepsilon}]
\end{aligned}
$$

which, by Jensen's inequality, is less than

$$
2 \cdot \sqrt[4]{0.5 \cdot 2 \cdot 8.5}=2 \cdot \sqrt[4]{8.5}=f(-6.5)
$$

for $\varepsilon>0$. This only shows that the function $f(x)$ takes its maximum at $x=-6.5$, not that it takes every other value only once in the intervals $[-15 ; 6.5[$ and $[-6.5 ; 2]$. If we look at the graph of $g(x)=\sqrt[4]{x}$, it is visible that its growth slows down as it moves to the right, from which it follows that for $\delta>\varepsilon$,

$$
f(-6.5+\delta)<f(-6.5+\varepsilon)
$$

$(\sqrt[4]{8.5-\delta}+\sqrt[4]{8.5+\delta}<\sqrt[4]{8.5-\varepsilon}+\sqrt[4]{8.5+\varepsilon}, \sqrt[4]{8.5+\delta}-\sqrt[4]{8.5+\varepsilon}<\sqrt[4]{8.5-\varepsilon}-\sqrt[4]{8.5-\delta})$, which can be illustrated similarly to the two segments shown in the graph.

![](https://cdn.mathpix.com/cropped/2024_05_02_67bca1062d9e2c6bbf7cg-2.jpg?height=318&width=482&top_left_y=1760&top_left_x=816)

A precise proof can also be given using the differentiation used in the III. solution.