1Cademy - Solving $$\sqrt{r+4}

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How to Solve a Radical Equation

Example

Solving $\sqrt{r+4} - r + 2 = 0$

Solve the radical equation $\sqrt{r + 4} - r + 2 = 0$ using the four-step procedure. This equation produces a quadratic after squaring, and one of the two candidate solutions turns out to be extraneous.

Step 1 — Isolate the radical. Add $r$ and subtract $2$ from both sides (equivalently, move $-r + 2$ to the right side):

$\sqrt{r + 4} = r - 2$

Step 2 — Square both sides. The left side simplifies via the Squaring Property. The right side is a binomial that must be expanded using the Binomial Squares Pattern $(a - b)^2 = a^2 - 2ab + b^2$ :

$(\sqrt{r + 4})^2 = (r - 2)^2$

$r + 4 = r^2 - 4r + 4$

Step 3 — Solve the new equation. Rearrange to standard form by moving all terms to one side:

$0 = r^2 - 5r$

Factor the right side: $0 = r(r - 5)$ . Apply the Zero Product Property:

$r = 0 \quad \text{or} \quad r - 5 = 0$

$r = 0 \quad \text{or} \quad r = 5$

Step 4 — Check both answers by substituting into the original equation:

For $r = 0$ : $\sqrt{0 + 4} - 0 + 2 = \sqrt{4} + 2 = 2 + 2 = 4 \neq 0$ . This is not a solution — $r = 0$ is an extraneous solution.

For $r = 5$ : $\sqrt{5 + 4} - 5 + 2 = \sqrt{9} - 3 = 3 - 3 = 0$ . Since $0 = 0$ is true, $r = 5$ is a solution. ✓

The solution is $r = 5$ . The value $r = 0$ is extraneous — it was introduced by the squaring step. When $r = 0$ , isolating the radical gives $\sqrt{4} = 0 - 2 = -2$ , which requires the principal square root to equal a negative number, an impossibility. This example reinforces that checking every candidate solution is essential, especially when squaring produces a quadratic with multiple solutions.

Updated 2026-04-21

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OpenStax Elementary Algebra

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