1Cademy - Solving $$y^2 - 6y = 16$$ by Completing the Square

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Completing the Square

Example

Solving $y^2 - 6y = 16$ by Completing the Square

Solve $y^2 - 6y = 16$ by completing the square, demonstrating the procedure when the linear coefficient is negative.

Step 1 — Isolate the variable terms. The variable terms are already on the left side.

Step 2 — Find $\left(\frac{1}{2} \cdot b\right)^2$ and add to both sides. The coefficient of $y$ is $-6$ , so $b = -6$ . Compute: $\left(\frac{1}{2}(-6)\right)^2 = (-3)^2 = 9$ . Add $9$ to both sides:

$y^2 - 6y + 9 = 16 + 9$

Step 3 — Factor the perfect square trinomial. Because the linear term is negative, the binomial square uses subtraction:

$(y - 3)^2 = 25$

Step 4 — Apply the Square Root Property:

$y - 3 = \pm\sqrt{25}$

Step 5 — Simplify and solve. Since $25$ is a perfect square ( $5^2 = 25$ ):

$y - 3 = \pm 5$

Write as two equations:

$y - 3 = 5 \implies y = 8$

$y - 3 = -5 \implies y = -2$

Step 6 — Check both solutions:

For $y = 8$ : $8^2 - 6(8) = 64 - 48 = 16$ ✓

For $y = -2$ : $(-2)^2 - 6(-2) = 4 + 12 = 16$ ✓

The solutions are $y = 8$ and $y = -2$ . This example shows how completing the square works when the linear coefficient is negative: halving $-6$ gives $-3$ , and squaring $-3$ still yields a positive constant ( $9$ ). The negative sign in the original expression determines that the factored form uses subtraction — $(y - 3)^2$ rather than $(y + 3)^2$ .

Updated 2026-04-21

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OpenStax Elementary Algebra

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