Solve the equation \[ \sqrt{x+4} + \sqrt{x+10} = \sqrt{8x + 18}. \]
By squaring both sides and rearranging, the equation becomes
\[\begin{align*} \left( \sqrt{x+4} + \sqrt{x+10} \right)^2 = \left( \sqrt{8x + 18} \right)^2 &\implies \left( \sqrt{x+4}\right)^2 + 2\sqrt{x+4}\sqrt{x+10} + \left( \sqrt{x+10} \right)^2 = \left( \sqrt{8x + 18} \right)^2 \\ &\implies(x+4) + 2\sqrt{x+4}\sqrt{x+10} + (x+10) = 8x + 18 \\ &\implies 2\sqrt{x+4}\sqrt{x+10} = 6x + 4 \\ &\implies \sqrt{x+4}\sqrt{x+10} = 3x + 2. \end{align*}\]Again squaring both sides, we have
\[\begin{align*} \left( \sqrt{x+4}\sqrt{x+10} \right)^2 = \left(3x + 2\right)^2 &\implies (x+4)(x+10) = (3x + 2)^2 \\ &\implies x^2 + 14x + 40 = 9x^2 + 12x + 4 \\ &\implies 8x^2 - 2x - 36 = 0 \\ &\implies 4x^2 - x - 18 = 0 \\ &\implies (4x - 9)(x + 2) = 0. \end{align*}\]The solutions to this final equation are
\[ x = \dfrac{9}{4} \quad\text{and}\quad x = -2. \]
But we need to be cautious here; our squaring may have introduced “false” solutions.
For example, the equation \(x+1 = 2\) clearly has the unique solution \(x = 1\). But if we square both sides, we get \((x+1)^2 = 4\), which has the solution \(x = 1\) and the “false” solution \(x = -3\).
So we need to check our answers by substituting back into our original equation.
If we do this, we see that \(\dfrac{9}{4}\) is a solution, but \(-2\) is not.
