Solve the equation $\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$.

Question

Solve the equation: $$\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$$

Here's what I've done.

Let $\sqrt{x + 2} = a$ and $\sqrt{3 - x} = b$

$\implies \left\{ \begin{align} a^2 + b^2 &= 5\\ a^2 - b^2 &= 2x - 1 \end{align} \right.$.

We have that $a - b = (x - 3)^2 \implies a + b = \dfrac{a^2 - b^2}{a + b} = \dfrac{2x - 1}{(x - 3)^2}$.

$\left\{ \begin{align} a = \dfrac{(a + b) + (a - b)}{2} = \dfrac{x^4 - 12x^3 + 54x^2 - 106x + 80}{2(x - 3)^2}\\ b = \dfrac{(a + b) - (a - b)}{2} = \dfrac{x^4 - 12x^3 + 54x^2 - 110x + 82}{2(x - 3)^2} \end{align} \right.$

Answer 1

$$\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9 \implies (\sqrt{x + 2} - 2) - (\sqrt{3 - x} - 1) = x^2 - 6x + 8$$

$$ \implies (x - 2)\left(\frac{1}{\sqrt{x + 2} + 2} + \frac{1}{\sqrt{3 - x} + 1}\right) = (x - 2)(x - 4)$$

$$\implies \left[ \begin{align*} x - 2 &= 0\\ \frac{1}{\sqrt{x + 2} + 2} + \frac{1}{\sqrt{3 - x} + 1} &= x - 4 \end{align*} \right.$$

For the second equality, we have that $-2 \le x \le 3$ and $x \ge 4$ as the needed condition for the left and right side of the equality. So there're not any solutions for the second equality.

That means $x - 2 = 0 \implies x = 2$.

Answer 2

$ t = \sqrt{3-x}\geq 0$, then we get $$\sqrt{5-t^2}-t=t^4$$

so $$5-t^2 = t^2+2t^5+t^8$$

so we have $$t^8+2t^5+2t^2-5=0$$

Since $f(t)=t^8+2t^5+2t^2$ is strictly increasing for positive $t$, given equation has at most one real (positive) solution and that is $t=1$, or $x= 2$

Answer 3

After squaring one times we obtain $$-2\sqrt{x+2}\sqrt{3-x}=(x^2-6x+9)^2-5$$ squaring again we obtain $$ \left( x-2 \right) \left( {x}^{7}-22\,{x}^{6}+208\,{x}^{5}-1096\,{x} ^{4}+3468\,{x}^{3}-6552\,{x}^{2}+6772\,x-2876 \right) =0$$

Answer 4

Let $3-x=t^2$, with $t>0$. Then

$$\sqrt{5-t^2}-t=t^4$$ and

$$p(t):=(t^4+t)^2-\left(\sqrt{5-t^2}\right)^2=t^8+2t^5+2t^2-5=0.$$

Now the derivative

$$8t^7+10t^4+4t$$ cancels for $t=0$ (minimum), and $8(t^3)^2+10t^3+4$ has no real root. As $p(0)<0$, the polynomial has exactly one positive and one negative root, which we discard.

Finally, by inspection, $t=1$ is a root and $x=2$ is the only solution.