Can we solve $\sqrt{4x+13}-\sqrt{x+1}=\sqrt{12-x}?$

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Squaring both sides of the equation gives us \[\begin{align*} &(4x+13)+(x+1)-2\sqrt{(4x+13)(x+1)} = 12-x\\ \Longrightarrow\quad& 5x+14-12+x = 2\sqrt{(4x+13)(x+1)}\\ \Longrightarrow\quad& 3x+1 = \sqrt{(4x+13)(x+1)}. \end{align*}\] Squaring again implies that \[\begin{align*} &9x^2+6x+1 = (4x+13)(x+1)=4x^2+17x+13 \\ \Longrightarrow\quad& 5x^2-11x-12 = 0 \\ \Longrightarrow\quad& (5x+4)(x-3) = 0 \end{align*}\]

and so \(x = -\dfrac{4}{5}\) or \(x = 3\).

We know now that these are the only two possibilities for a solution. There is a problem, however; squaring may have intrduced false roots.

So we need to check whether or not the possible solutions above satisfy our original equation.

If \(x = -\dfrac{4}{5}\), then the left-hand side is \[\sqrt{4\left(-\dfrac{4}{5}\right)+13}-\sqrt{-\dfrac{4}{5} + 1} = \sqrt{\dfrac{49}{5}}-\sqrt{\dfrac{1}{5}}=\dfrac{1}{\sqrt{5}}(7-1)=\dfrac{6}{\sqrt{5}},\] while the right-hand side is \[\sqrt{12+\dfrac{4}{5}}=\sqrt{\dfrac{64}{5}}=\dfrac{8}{\sqrt{5}},\] which are not equal. Thus \(x=-\dfrac{4}{5}\) is not a solution.

If \(x=3\), then the left-hand side is \[\sqrt{4\times 3+13}-\sqrt{3 + 1}=5-2=3\] while the right-hand side is \[\sqrt{12-3}=\sqrt{9}=3.\] Thus \(x=3\) is the only solution to the original equation.

Let \(f(x)=3x^3+14x^2+2x-4\). Assume the given rational root is \(\dfrac{a}{b}\), where the highest common factor of \(a\) and \(b\) is \(1\) and, without loss of generality, \(b>0\).

Substituting this solution into \(f(x)=0\) and multiplying through by \(b^3\) we find \[3a^3+14a^2b+2ab^2-4b^3=0.\] Note that this means \(b\) divides \(3a^3\) (since it divides every other term) and so \(b\) is \(1\) or \(3\) (since it cannot share a factor with \(a\)).

Now \(2\) divides \(14a^2b+2ab^2-4b^3\), and so \(2\) divides \(3a^3\), and thus \(2\) divides \(a\), say \(a = 2c\). This gives us \[6c^3 + 14c^2b + cb^3 - b^3 = 0,\] and so \(c\) divides \(b^3\). But \(c\) and \(b\) have no common factor, and so \(c\) must be \(\pm 1\), and \(a\) must be \(\pm 2\).

So the only possible rational roots for the equation are \(\dfrac{2}{3}, -\dfrac{2}{3}, 2\) or \(-2\).

It is easy to check that \(-\dfrac{2}{3}\) is the only one to satisfy the equation. Thus \(x=-\dfrac{2}{3}\) is the rational root. Factorising the cubic we find \[3x^3 + 14x^2 + 2x - 4 = (3x+2)(x^2+4x-2)=0.\] So either \(x = -\dfrac{2}{3}\), or \(x^2+4x-2=0\) and so \[x=\dfrac{-4\pm\sqrt{16-4\times(-2)}}{2}=-2\pm\sqrt{6}.\]