1Cademy - Solving $$(y - 7)^2 = 12$$ Using the Square Root Property

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Square Root Property

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Solving $(y - 7)^2 = 12$ Using the Square Root Property

Solve $(y - 7)^2 = 12$ by applying the Square Root Property to a squared binomial. Unlike $(x - 3)^2 = 16$ where the constant was a perfect square, here $12$ is not a perfect square, so the solutions involve simplified radicals.

Apply the Square Root Property:

$y - 7 = \pm\sqrt{12}$

Simplify the radical. The largest perfect square factor of $12$ is $4$ . Apply the Product Property:

$\sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3}$

So $y - 7 = \pm 2\sqrt{3}$ .

Solve for $y$ by adding $7$ to both sides:

$y = 7 \pm 2\sqrt{3}$

Rewrite as two solutions:

$y = 7 + 2\sqrt{3}$ or $y = 7 - 2\sqrt{3}$

Check both solutions by substituting into the original equation:

For $y = 7 + 2\sqrt{3}$ : $(7 + 2\sqrt{3} - 7)^2 = (2\sqrt{3})^2 = 4 \cdot 3 = 12$ ✓

For $y = 7 - 2\sqrt{3}$ : $(7 - 2\sqrt{3} - 7)^2 = (-2\sqrt{3})^2 = 4 \cdot 3 = 12$ ✓

The solutions are $y = 7 + 2\sqrt{3}$ and $y = 7 - 2\sqrt{3}$ . When the constant on the right side is not a perfect square, the radical must be simplified using the Product Property of Square Roots. The solutions are then expressed as a rational number combined with a simplified radical term.

Updated 2026-04-21

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