how can I solve the following equation (without complex numbers)? $$
\sqrt{45-x^2} = 3- x^2
$$
$$
\sqrt{13-x^2} = 7 - x^2
$$
I have tried the Quadratic formula.  I always came up with the solution of $\{-3, 3\}$ by the first one and $\{-3, 3\}$ by the second one which is both wrong. Does this happen because I have to pay attention to a rule when substituting in a root, which I don't know, or how should I arrive at the solution that the first has none according to the graph and the second has the solution set according to the graph $\{-2,2\}$? I search a lot in the internet but there was no solution.
 A: Consider the first equation:
A general strategy would be to square both sides:
$$\sqrt{45-x^2} = 3- x^2 \implies 45-x^2 = (3-x^2)^2 = 9+x^4-6x^2 \iff x^4-5x^2-36=0$$
Using $a=x^2$, the equation becomes $a^2-5a-36=0$ with $a\ge 0$. The Quadratic Formula indeed say that the solutions are $a=9$ and $a=-4$. Since $a\ge 0$, only $a=9$ is solution, so $x=\pm 3$.
But! The first step of computation ($\sqrt{45-x^2}=3-x^2 \implies 45-x^2=(3-x^2)^2$) is an implication, not an equivalence, so it is important to check that $3$ and $-3$ are indeed solution. As you check, they are not, so the equation has no real solution.
What are $3$ and $-3$ doing here? The equation $45-x^2=(3-x^2)^2$ is actually equivalent to $3-x^2=\pm\sqrt{45-x^2}$ (assuming $45-x^2\ge 0$). $4$ and $-3$ are solutions of the "twin" equation $3-x^2=-\sqrt{45-x^2}$.

Other method to see this equation has no solution: $3-x^2=\sqrt{45-x^2}$ implies that $3-x^2\ge 0$ so $x^2\le 3$, hence $3-x^2\le 3$. This in turn implies that $45-x^2\ge 42$ so $\sqrt{45-x^2}\ge\sqrt{42}>3\ge 3-x^2$ so the equation has no real solution.
A: Here my attempt,
$\sqrt{\mathrm{45}−{x}^{\mathrm{2}} }=\mathrm{3}−{x}^{\mathrm{2}}  \\ $
$\sqrt{\mathrm{45}−{x}^{\mathrm{2}} }=\left(\mathrm{45}−{x}^{\mathrm{2}} \right)−\mathrm{42} \\ $
$\left(\sqrt{\mathrm{45}−{x}^{\mathrm{2}} }\right)^{\mathrm{2}} −\sqrt{\mathrm{45}−{x}^{\mathrm{2}} }−\mathrm{42}=\mathrm{0} \\ $
$\left(\sqrt{\mathrm{45}−{x}^{\mathrm{2}} }+\mathrm{6}\right)\left(\sqrt{\mathrm{45}−{x}^{\mathrm{2}} }−\mathrm{7}\right)=\mathrm{0} \\ $
$\sqrt{\mathrm{45}−{x}^{\mathrm{2}} }+\mathrm{6}=\mathrm{0}\Rightarrow{No}\:{Solution} \\ $
$\mathrm{45}−{x}^{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} \Rightarrow{x}^{\mathrm{2}} =-\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $
${x}=\pm\mathrm{2}{i}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $
$.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $
$\sqrt{\mathrm{13}−{x}^{\mathrm{2}} }=\mathrm{7}−{x}^{\mathrm{2}}  \\ $
$\sqrt{\mathrm{13}−{x}^{\mathrm{2}} }=\left(\mathrm{13}−{x}^{\mathrm{2}} \right)−\mathrm{6} \\ $
$\left(\sqrt{\mathrm{13}−{x}^{\mathrm{2}} }\right)^{\mathrm{2}} −\sqrt{\mathrm{13}−{x}^{\mathrm{2}} }−\mathrm{6}=\mathrm{0} \\ $
$\left(\sqrt{\mathrm{13}−{x}^{\mathrm{2}} }+\mathrm{2}\right)\left(\sqrt{\mathrm{13}−{x}^{\mathrm{2}} }−\mathrm{3}\right)=\mathrm{0} \\ $
$\sqrt{\mathrm{13}−{x}^{\mathrm{2}} }+\mathrm{2}=\mathrm{0}\Rightarrow{No}\:{Solution} \\ $
$\mathrm{13}−{x}^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} \Rightarrow{x}^{\mathrm{2}} =\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $
${x}=\pm\mathrm{2} \\ $
Hope acceptable.
A: Define $u=3-x^2$. We have
$$\sqrt{42+u}=u$$
$$u^2=42+u$$
$$(u-7)(u+6)=0$$
$u=-6$ is extranneous. So $u=7$ is the only potential solution. This means that $x^2=-4$, which clearly has no real solutions.
For the other system, define $u=7-x^2$. We have
$$\sqrt{6+u}=u$$
$$u^2-u-6=0$$
$$(u-3)(u+2)=0$$
$u=-2$ is extranneous. So $u=3$ is the only potential solution. This means that $x^2=4$, so the only solutions are $x=\pm 2$.
