1Cademy - Solving $$\sqrt{p-1} + 1 = p$$

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How to Solve a Radical Equation

Example

Solving $\sqrt{p-1} + 1 = p$

Solve the radical equation $\sqrt{p - 1} + 1 = p$ using the four-step procedure. This equation is more complex than earlier examples because squaring both sides produces a quadratic equation that must be solved by factoring.

Step 1 — Isolate the radical. Subtract $1$ from both sides:

$\sqrt{p - 1} + 1 - 1 = p - 1$

$\sqrt{p - 1} = p - 1$

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 $(a - b)^2 = a^2 - 2ab + b^2$ :

$(\sqrt{p - 1})^2 = (p - 1)^2$

$p - 1 = p^2 - 2p + 1$

Step 3 — Solve the new equation. The result is a quadratic equation. Move all terms to one side to obtain standard form:

$0 = p^2 - 3p + 2$

Factor the right side: $0 = (p - 1)(p - 2)$ . Apply the Zero Product Property:

$p - 1 = 0 \quad \text{or} \quad p - 2 = 0$

$p = 1 \quad \text{or} \quad p = 2$

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

For $p = 1$ : $\sqrt{1 - 1} + 1 = \sqrt{0} + 1 = 0 + 1 = 1$ , and the right side is $1$ . Since $1 = 1$ is true, $p = 1$ is a solution. ✓

For $p = 2$ : $\sqrt{2 - 1} + 1 = \sqrt{1} + 1 = 1 + 1 = 2$ , and the right side is $2$ . Since $2 = 2$ is true, $p = 2$ is a solution. ✓

The solutions are $p = 1$ and $p = 2$ . Unlike earlier radical equation examples where squaring produced a linear equation, here the non-radical side $(p - 1)$ is a binomial that produces a trinomial $p^2 - 2p + 1$ when squared. Correctly applying the Binomial Squares Pattern — including the middle term $-2p$ — is essential. Omitting the middle term is a common error.

Updated 2026-04-21

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OpenStax Elementary Algebra

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