How do we solve an equation containing three square roots?

How do we solve an equation containing three square roots?

Question

Solve the equation $\sqrt{x+4} + \sqrt{x+10} = \sqrt{8x + 18}.$

Suggestion

Do we know a way of ‘undoing’ square roots?

What happens if we try squaring both sides? Can things be simplified?

A further suggestion

Solution

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.