<s>[INST] Easy way of tackling $\sqrt{x^2}+x=\sqrt{5}$ Solve for x
$$\sqrt{x^2}+x=\sqrt{5}\tag1$$
$$\sin(\sqrt{x^2}+x)=\sin({\sqrt{5}})\tag2$$
$$\sin(\sqrt{x^2})\cos(x)+\cos(\sqrt{x^2})\sin(x)=\sin(\sqrt{5})\tag3$$
$$\sin(\sqrt{x^2})+\cos(\sqrt{x^2})\tan(x)=\sin(\sqrt{5})\sec(x)\tag4$$
$$\sin(\sqrt{x^2})+\cos(\sqrt{x^2})\sqrt{\sec^2(x)-1}=\sin(\sqrt{5})\sec(x)\tag5$$
finally after all the hard work $(5)$ reveals
$$x+x=\sqrt{5}\tag6$$
$$x=\frac{\sqrt{5}}{2}\tag7$$
Is there another easy way solving this $(1)?$ 
 [/INST] There is another way to solve this.
Using the fact that $\sqrt{x^2} = |x|$. Your equation turns out to be $$|x|+x=\sqrt{5}$$
First look for solutions where $x\geq 0$ in this case you get $x+x=\sqrt{5}$ and so $x=\frac{\sqrt{5}}{2}$.
Now if $x<0$ then $|x|=-x$ and so you get $-x+x=\sqrt{5}$ and clearly no $x$ satisfies this condition.
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