1Cademy - Solving $$\sqrt{q - 2} + 3 = \sqrt{4q + 1}$$

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Repeating the Isolate-and-Square Steps in Radical Equations

Example

Solving $\sqrt{q - 2} + 3 = \sqrt{4q + 1}$

Solve the radical equation $\sqrt{q - 2} + 3 = \sqrt{4q + 1}$ , which requires squaring twice and produces a quadratic equation that must be solved by factoring.

Step 1 — Identify the isolated radical. The right side is a single square root $\sqrt{4q + 1}$ . Square both sides:

$(\sqrt{q - 2} + 3)^2 = (\sqrt{4q + 1})^2$

Step 2 — Expand using the Binomial Squares Pattern. With $a = \sqrt{q - 2}$ and $b = 3$ , apply $(a + b)^2 = a^2 + 2ab + b^2$ :

$q - 2 + 6\sqrt{q - 2} + 9 = 4q + 1$

A radical $6\sqrt{q - 2}$ remains, so the procedure must be repeated.

Step 3 — Isolate the remaining radical. Combine constants on the left ( $-2 + 9 = 7$ ) and rearrange:

$6\sqrt{q - 2} = 3q - 6$

Step 4 — Square both sides again:

$(6\sqrt{q - 2})^2 = (3q - 6)^2$

$36(q - 2) = 9q^2 - 36q + 36$

Step 5 — Distribute and solve the quadratic. Distribute the left side: $36q - 72 = 9q^2 - 36q + 36$ . Move all terms to one side:

$0 = 9q^2 - 72q + 108$

Factor out $9$ : $0 = 9(q^2 - 8q + 12)$ . Factor the trinomial: $0 = 9(q - 6)(q - 2)$ . Apply the Zero Product Property:

$q = 6 \quad \text{or} \quad q = 2$

Step 6 — Check. Both candidate solutions satisfy the original equation (verification is left to the reader).

The solutions are $q = 6$ and $q = 2$ . Unlike Example 9.83 where the second squaring produced a linear equation, this example produces a quadratic because the right side of the isolated-radical equation ( $3q - 6$ ) is a binomial in $q$ rather than a constant. Squaring that binomial generates a $q^2$ term, requiring factoring and the Zero Product Property to find the solutions.

Updated 2026-04-21

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OpenStax Elementary Algebra

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