If $a^2+a b+b^2=40$ and $a^2-\sqrt{a b}+b=5$, then find $a^2+\sqrt{a b}+b$ I was given this problem to solve with elementary methods (High School level).

Knowing that
$$\begin{align}
a^2+a b+b^2 &=40 \\
a^2-\sqrt{a b}+b &=\phantom{0}5
\end{align}$$
find
$$a^2+\sqrt{a b}+b$$

I tried to look for $\sqrt{a b}$, since the requested quantity is
$$a^2+\sqrt{a b}+b=(a^2-\sqrt{a b}+b)+2\sqrt{ab}=5 + 2\sqrt{ab}$$
So I set
$$x=a^2;\;y=\sqrt{ab};\;b=z$$
and the system became
$$x+y^2+z^2=40;\;x-y-z=5$$
subtracting the two equations I got
$$z^2-z+y^2+y-35=0$$
which has one real solution when the discriminant is zero.
That is
$1-4(y^2+y-35)=0$
and then
$y=\frac{1}{2} \left(-1\pm\sqrt{142}\right)$
and finally
$$a^2+\sqrt{a b}+b=4\pm\sqrt{142}$$
I know that there are other solutions because I've found them with Wolfram Mathematica, but I couldn't find them with elementary methods.
Any help will be appreciated.

One way to do this is to graph out these equations of lines.
Find the values of $a,b$ and plug into the desired output equation:

*

*Equation of a line:
see link here.


*Look at the possible values and


*plug answers into the final equation:

