Review question

# How do we solve an equation containing three square roots? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5847

## Solution

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.