Calculate the number of solutions Calculate the number of solutions presented by the equation.
$$\sqrt{1-x}+\sqrt{1-2x}+\sqrt{1-4x}=x^2+2$$
What I thought: The LHS is concave, the RHS is convex
 A: Hints:


*

*Your observation "The LHS is concave, the RHS is convex" puts an upper bound on the number of solutions

*The square roots on the LHS require $x \le \frac14$ 

*Consider the difference between the two sides for some sensible values of $x$: $\frac14$ and $0$ look obvious possibilities and you might also think about a very negative value 

A: Hint : If $X= \sqrt{a}+ \sqrt{b}+\sqrt{c}$ ... keep squaring and rearranging ...
\begin{eqnarray*}
X &=& \sqrt{a}+ \sqrt{b}+\sqrt{c} \\
\frac{X^2-a-b-c}{2} &=& \sqrt{ab}+ \sqrt{bc}+\sqrt{ca} \\
\left(\frac{X^2-a-b-c}{2} \right)^2 &=& 2\sqrt{abc}(\sqrt{a}+ \sqrt{b}+\sqrt{c})=2\sqrt{abc}X. \\
\end{eqnarray*}
Now square one final time & what order is the polynomial ?
A: I think the argumentation becomes easier if you transform the given equation to: 
$$x+\frac{1}{1+\sqrt{1-x}}+\frac{2}{1+\sqrt{1-2x}}+\frac{4}{1+\sqrt{1-4x}}=\frac{1}{x}$$
The left side is almost linear (beside a small rest), growing from the $3^{rd}$ till to the $1^{st}$ part of the coordinate plane. The right side has one falling curve in the $3^{rd}$ part and one falling curve in the $1^{st}$ part. So there is one negative (we can see this using $x\to -\infty$ and $x=-2$) and one positive solution (we can see this using $x\downarrow 0$ and $x=1/4$) .
A: You can use the Intermediate Value Theorem to prove the equation has a non null solution on $\mathbb R_0^-$ and another non null solution on $[0,\frac{1}{4}]$ using each equation member as a function itself. Then you can use the fact that each of those functions are injective on each respective interval to prove that each one of those solutions is unique.
Therefore, the equation has two solutions (as you can see next):

