Solve this

Given :

$4^{(x+1)}+4^{(1-x)}=10$

$\Rightarrow 4^{x} \cdot 4+4^{1} \cdot \frac{1}{4^{x}}=10$

Let $4^{x}$ be $y$.

$\therefore 4 y+\frac{4}{y}=10$

$\Rightarrow 4 y^{2}-10 y+4=0$

$\Rightarrow 4 y^{2}-8 y-2 y+4=0$

$\Rightarrow 4 y(y-2)-2(y-2)=0$

$\Rightarrow y=2$ or $y=\frac{2}{4}=\frac{1}{2}$

$\Rightarrow 4^{x}=2$ or $\frac{1}{2}$

$\Rightarrow 4^{x}=2^{2 x}=2^{1}$ or $2^{2 x}=2^{-1}$

$\Rightarrow x=\frac{1}{2}$ or $x=\frac{-1}{2}$

Hence, $\frac{1}{2}$ and $\frac{-1}{2}$ are roots of the given equation.