Solve over the real numbers: $$(x^2+x+1)^{1/3}+(2x+1)^{1/2}=2$$
I know for the second radical to be defined $x≥-0,5$ and I've attempted various methods I've solved other such equations with but to no avail; if I could write $x^2+x-7$ in terms of $2x+1$ to use a convenient notation in $x^2+x-7-6(2x+1)+12(2x+1)^{1/2}+(2x+1)(2x+1)^{1/2}=0$
I think I could solve it. I've ran it through Wolfram and the only real solution is $0$ however how the conclusion was reached I am not aware.
On
$f(x) = ((x+\frac 12)^2 + \frac 34)^\frac 13 + (2(x+\frac 12))^\frac 12$
The second term is only defined if $x\ge - \frac 12$
Both terms are strictly increasing if $x>\frac 12$
There is only one solution.
And by inspection if $x = 0$
$f(0) = 1^\frac 13 + 1^\frac 12 = 2$
On
If we put $$a =(x^2+x+1)^{1/3}\;\;\;{\rm and}\;\;\;b= (2x+1)^{1/2}$$
So $$a^3= x^2+x+1 \;\;\;{\rm and}\;\;\;b^2= 2x+1\;\;\;{\rm and}\;\;\; a+b=2$$
and thus $$4a^3 = 4x^2+4x+4 = (2x+1)^2+3 = b^4+3$$
and finally $$ 4(2-b)^3 = b^4+3$$ ...
$$ b^4+4b^3-24b^2+48b-29=0$$
$$ (b-1)(b^3+ \underbrace{ 5b^2-19b+29}_{f(b)})=0$$
Since the discriminat of $f$ is negative $f$ is always positive. Now $b\geq 0$ so $b^3+f(b)>0$ and thus $b=1$ is only solution i.e. $x=0$.
Let $f(x)=\sqrt[3]{x^2+x+1}+\sqrt{2x+1}$, then $$f'(x)=\frac{2x+1}{3\sqrt[3]{(x^2+x+1)^2}}+\frac{1}{\sqrt{2x+1}}$$ If there is no solution to $f'(x)=0$ then $f(x)=2$ has atmost one solution. $$f'(x)=0\\\frac{(2x+1)\sqrt{2x+1}}{3\sqrt[3]{(x^2+x+1)^2}}=-1$$
This last equation is impossible, since the left hand side is always non-negative. So there is only one solution to $f(x)=2$, which you have found.